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7.5: Loans - Mathematics


In the last section, you learned about payout annuities.

In this section, you will learn about conventional loans (also called amortized loans or installment loans). Examples include auto loans and home mortgages. These techniques do not apply to payday loans, add-on loans, or other loan types where the interest is calculated up front.

One great thing about loans is that they use exactly the same formula as a payout annuity. To see why, imagine that you had $10,000 invested at a bank, and started taking out payments while earning interest as part of a payout annuity, and after 5 years your balance was zero. Flip that around, and imagine that you are acting as the bank, and a car lender is acting as you. The car lender invests $10,000 in you. Since you’re acting as the bank, you pay interest. The car lender takes payments until the balance is zero.

Loans Formula

P0 is the balance in the account at the beginning (the principal, or amount of the loan).

d is your loan payment (your monthly payment, annual payment, etc)

r is the annual interest rate in decimal form.

k is the number of compounding periods in one year.

N is the length of the loan, in years

Like before, the compounding frequency is not always explicitly given, but is determined by how often you make payments.

When do you use this

The loan formula assumes that you make loan payments on a regular schedule (every month, year, quarter, etc.) and are paying interest on the loan.

Compound interest: One deposit

Annuity: Many deposits.

Payout Annuity: Many withdrawals

Loans: Many payments

Example 11

You can afford $200 per month as a car payment. If you can get an auto loan at 3% interest for 60 months (5 years), how expensive of a car can you afford? In other words, what amount loan can you pay off with $200 per month?

In this example,

d = $200 the monthly loan payment

r = 0.03 3% annual rate

k = 12 since we’re doing monthly payments, we’ll compound monthly

N = 5 since we’re making monthly payments for 5 years

We’re looking for P0, the starting amount of the loan.

You can afford a $11,120 loan.

You will pay a total of $12,000 ($200 per month for 60 months) to the loan company. The difference between the amount you pay and the amount of the loan is the interest paid. In this case, you’re paying $12,000-$11,120 = $880 interest total.

Example 12

You want to take out a $140,000 mortgage (home loan). The interest rate on the loan is 6%, and the loan is for 30 years. How much will your monthly payments be?

In this example,

We’re looking for d.

r = 0.06 6% annual rate

k = 12 since we’re paying monthly

N = 30 30 years

P0 = $140,000 the starting loan amount

In this case, we’re going to have to set up the equation, and solve for d.

You will make payments of $839.37 per month for 30 years.

You’re paying a total of $302,173.20 to the loan company: $839.37 per month for 360 months. You are paying a total of $302,173.20 – $140,000 = $162,173.20 in interest over the life of the loan.

Try it Now 4

Janine bought $3,000 of new furniture on credit. Because her credit score isn’t very good, the store is charging her a fairly high interest rate on the loan: 16%. If she agreed to pay off the furniture over 2 years, how much will she have to pay each month?


Auto Loan Calculator

The Auto Loan Calculator is mainly intended for car purchases within the U.S. People outside the U.S. may still use the calculator, but please adjust accordingly. If only the monthly payment for any auto loan is given, use the Monthly Payments tab (reverse auto loan) to calculate the actual vehicle purchase price and other auto loan information.

Auto Loans

Most people turn to auto loans during vehicle purchase. They work as any generic, secured loan from a financial institution does with a typical term of 36, 60, 72, or 84 months in the U.S. Each month, repayment of principal and interest must be made from borrowers to auto loan lenders. Money borrowed from a lender that isn't paid back can result in the car being legally repossessed.

Dealership Financing vs. Direct Lending

Generally, there are two main financing options available when it comes to auto loans: direct lending or dealership financing. With the former, it comes in the form of a typical loan originating from a bank, credit union, or financial institution. Once a contract has been entered with a car dealer to buy a vehicle, the loan is used from the direct lender to pay for the new car. Dealership financing is somewhat similar except that the auto loan, and thus paperwork, is initiated and completed through the dealership instead. Auto loans via dealers are usually serviced by captive lenders that are often associated with each car make. The contract is retained by the dealer, but is often sold to a bank or other financial institution called an assignee that ultimately services the loan.

Direct lending provides more leverage for buyers to walk into a car dealer with most of the financing done on their terms, as it places further stress on the car dealer to compete with a better rate. Getting pre-approved doesn't tie car buyers down to any one dealership, and their propensity to simply walk away is much higher. With dealer financing, the potential car buyer has fewer choices when it comes to rate shopping, though it's there for convenience for anyone who doesn't want to spend time shopping, or cannot get an auto loan through direct lending.

Often, to promote auto sales, car manufacturers offer good financing deals via dealers. Consumers in the market for a new car should start their search for financing with car manufacturers. It is not rare to get low interest rates like 0%, 0.9%, 1.9%, or 2.9% from car manufacturers.

Vehicle Rebates

Car manufacturers may offer vehicle rebates to further incentivize buyers. Depending on the state, the rebate may or may not be taxed accordingly. For example, purchasing a vehicle at $30,000 with a cash rebate of $2,000 will have sales tax calculated based on the original price of $30,000, not $28,000. Luckily, a good portion of states do not do this and don't tax cash rebates. They are Alaska, Arizona, Delaware, Iowa, Kansas, Kentucky, Louisiana, Massachusetts, Minnesota, Missouri, Montana, Nebraska, New Hampshire, Oklahoma, Oregon, Pennsylvania, Rhode Island, Texas, Utah, Vermont, and Wyoming.

Generally, rebates are only offered for new cars. While some used car dealers do offer cash rebates, this is rare due to the difficulty involved in determining the true value of the vehicle.

A car purchase comes with costs other than the purchase price, the majority of which are fees that can normally be rolled into the financing of the auto loan or paid upfront. However, car buyers with low credit scores might be forced into paying fees upfront. The following is a list of common fees associated with car purchases in the U.S.

  • Sales Tax&mdashMost states in the U.S. collect sales tax for auto purchases. It is possible to finance the cost of sales tax with the price of the car, depending on the state the car was purchased in. Alaska, Delaware, Montana, New Hampshire, and Oregon are the five states that don't charge sales tax.
  • Document Fees&mdashThis is a fee collected by the dealer for processing documents like title and registration.
  • Title and Registration Fees&mdashThis is the fee collected by states for vehicle title and registration.
  • Advertising Fees&mdashThis is a fee that the regional dealer pays for promoting the manufacturer's automobile in the dealer's area. If not charged separately, advertising fees are included in the auto price. A typical price tag for this fee is a few hundred dollars.
  • Destination Fee&mdashThis is a fee that covers the shipment of the vehicle from the plant to the dealer's office. This fee is usually between $900 and $1,500.
  • Insurance&mdashIn the U.S., auto insurance is strictly mandatory to be regarded as a legal driver on public roads and is usually required before dealers can process paperwork. When a car is purchased via loan and not cash, full coverage insurance is often mandatory. Auto insurance can possibly run more than $1,000 a year for full coverage. Most auto dealers can provide short-term (1 or 2 months) insurance for paper work processing so new car owners can deal with proper insurance later.

If the fees are bundled into the auto loan, remember to check the box 'Include All Fees in Loan' in the calculator. If they are paid upfront instead, leave it unchecked. Should an auto dealer package any mysterious special charges into a car purchase, it would be wise to demand justification and thorough explanations for their inclusion.

Auto Loan Strategies

Probably the most important strategy to get a great auto loan is to be well-prepared. This means determining what is affordable before heading to a dealership first. Knowing what kind of vehicle is desired will make it easier to research and find the best deals to suit your individual needs. Once a particular make and model is chosen, it is generally useful to have some typical going rates in mind to enable effective negotiations with a car salesman. This includes talking to more than one lender and getting quotes from several different places. Car dealers, like many businesses, want to make as much money as possible from a sale, but often, given enough negotiation, are willing to sell a car for significantly less than the price they initially offer. Getting a preapproval for an auto loan through direct lending can aid negotiations.

Credit, and to a lesser extent, income, generally determines approval for auto loans, whether through dealership financing or direct lending. In addition, borrowers with excellent credit will most likely receive lower interest rates, which will result in paying less for a car overall. Borrowers can improve their chances to negotiate the best deals by taking steps towards achieving better credit scores before taking out a loan to purchase a car.

Cash Back vs Low Interest

When purchasing a vehicle, many times auto manufacturers may offer either a cash vehicle rebate or a low-interest rate. A cash rebate instantly reduces the purchasing price of the car, but a lower rate can potentially result in savings in interest payments. The choice between the two will be different for everyone. For more information about or to do calculations involving this decision, please go to the Cash Back vs Low Interest Calculator.

Paying off an auto loan earlier than usual not only shortens the length of the loan, but can also result in interest savings. However, some lenders have early payoff penalty or terms restricting early payoff. It is important to examine the details carefully before signing an auto loan contract.

Consider Other Options

Although the allure of a new car can be strong, buying a pre-owned car even if only a few years removed from new can usually result in significant savings new cars depreciate as soon as they are driven off the lot, sometimes by more than 10% of their values this is called off-the-lot depreciation, and is an alternative option for prospective car buyers to consider.

People who just want a new car for the enjoyment of driving a new car may also consider a lease, which is, in essence, a long-term rental that normally costs less upfront than a full purchase. For more information about or to do calculations involving auto leases, please visit the Auto Lease Calculator.

In some cases, a car might not even be needed! If possible, consider public transportation, carpool with other people, bike, or walk instead.

Buying a Car with Cash Instead

Although most car purchases are done with auto loans in the U.S. there are benefits to buying a car outright with cash.

  • Avoid Monthly Payments&mdashPaying with cash relinquishes a person of the responsibility of making monthly payments. This can be a huge emotional benefit for anyone who would prefer not to have a large loan looming over their head for the next few years. In addition, the possibility of late fees for late monthly payments no longer exists.
  • Avoid Interest&mdashNo financing involved in the purchase of a car means there will be no interest charged, which will result in a lower overall cost to own the car. As a very simple example, borrowing $32,000 for five years at 6% will require a payment of $618.65 per month, with a total interest payment of $5,118.98 over the life of the loan. In this scenario, paying in cash will save $5,118.98.
  • Future Flexibility&mdashBecause ownership of a car is 100% and immediate after paying in full, there aren't any restrictions on the car, such as the right to sell it after several months, use less expensive insurance coverage, and make certain modifications to the car.
  • Avoid Overbuying&mdashPaying in full with a single amount will limit car buyers to what is within their immediate, calculated budget. On the other hand, financed purchases are less concrete, and have the potential to result in car buyers buying more than what they can afford long term it's easy to be tempted to add a few extra dollars to a monthly payment to stretch the loan length out for a more expensive car. To complicate matters, car salesmen tend to use tactics such as fees and intricate financing in order to get buyers to buy out of their realm. All of this can be avoided by paying in cash.
  • Discounts&mdashIn some cases, car purchases can come with the option of either an immediate rebate or low-interest financing. Certain rebates are only offered to cash purchases.
  • Avoid Underwater Loan&mdashWhen it comes to financing a depreciating asset, there is the chance that the loan goes underwater, which means more is owed on the asset than its current worth. Auto loans are no different, and paying in full completely avoids this scenario.

There are a lot of benefits to paying with cash for a car purchase, but that doesn't mean everyone should do it. Situations exist where financing with an auto loan can make more sense to a car buyer, even if they have enough saved funds to purchase the car in a single payment. For example, if a very low interest rate auto loan is offered on a car purchase and there exist other opportunities to make greater investments with the funds, it might be more worthwhile to invest the money instead to receive a higher return. Also, a car buyer striving to achieve a higher credit score can choose the financing option, and never miss a single monthly payment on their new car in order to build their scores, which aid other areas of personal finance. It is up to each individual to determine which the right decision is.

Trade-in Value

A trade-in is the process of selling your vehicle to the dealership in exchange for credit toward purchasing another vehicle. Don't expect too much value when trading in old cars to dealerships. Selling old cars privately and using the funds for a future car purchase tends to result in a more financially-desirable outcome.

In most of the states that collect sales tax on auto purchases (not all do), the sales tax collected is based on the difference between the new car and trade-in price. For a $30,000 new car purchase with a $10,000 trade-in value, the tax paid on the new purchase with an 8% tax rate is:

Some states do not offer any sales tax reduction with trade-ins, including California, District of Columbia, Hawaii, Kentucky, Maryland, Michigan, Montana, and Virginia. This Auto Loan Calculator automatically adjusts the method used to calculate sales tax involving Trade-in Value based on the state provided.

Using the values from the example above, if the new car was purchased in a state without a sales tax reduction for trade-ins, the sales tax would be:

This comes out to be an $800 difference which could be reason for people selling a car in these states to consider a private sale.


Calculating Interest on a One-Year Loan

If you borrow $1,000 from a bank for one year and have to pay $60 in interest for that year, your stated interest rate is 6%. Here's the calculation:

Effective Rate on a Simple Interest Loan = Interest/Principal = $60/$1,000 = 6%

Your annual percentage rate or APR is the same as the stated rate in this example because there is no compound interest to consider. This is a simple interest loan.

Meanwhile, this particular loan becomes less favorable if you keep the money for a shorter period of time. For example, if you borrow $1,000 from a bank for 120 days and the interest rate remains at 6%, the effective annual interest rate is much higher.

Effective rate = Interest/Principal X Days in the Year (360)/Days Loan Is Outstanding

Effective rate on a Loan with a Term of Less Than One Year = $60/$1,000 X 360/120 = 18%

The effective rate of interest is 18% since you only have use of the funds for 120 days instead of 360 days.


Contents

There is a unique complex Lie algebra of type E7, corresponding to a complex group of complex dimension 133. The complex adjoint Lie group E7 of complex dimension 133 can be considered as a simple real Lie group of real dimension 266. This has fundamental group Z/2Z, has maximal compact subgroup the compact form (see below) of E7, and has an outer automorphism group of order 2 generated by complex conjugation.

As well as the complex Lie group of type E7, there are four real forms of the Lie algebra, and correspondingly four real forms of the group with trivial center (all of which have an algebraic double cover, and three of which have further non-algebraic covers, giving further real forms), all of real dimension 133, as follows:

  • The compact form (which is usually the one meant if no other information is given), which has fundamental group Z/2Z and has trivial outer automorphism group.
  • The split form, EV (or E7(7)), which has maximal compact subgroup SU(8)/<±1>, fundamental group cyclic of order 4 and outer automorphism group of order 2.
  • EVI (or E7(-5)), which has maximal compact subgroup SU(2)·SO(12)/(center), fundamental group non-cyclic of order 4 and trivial outer automorphism group.
  • EVII (or E7(-25)), which has maximal compact subgroup SO(2)·E6/(center), infinite cyclic fundamental group and outer automorphism group of order 2.

For a complete list of real forms of simple Lie algebras, see the list of simple Lie groups.

The compact real form of E7 is the isometry group of the 64-dimensional exceptional compact Riemannian symmetric space EVI (in Cartan's classification). It is known informally as the "quateroctonionic projective plane" because it can be built using an algebra that is the tensor product of the quaternions and the octonions, and is also known as a Rosenfeld projective plane, though it does not obey the usual axioms of a projective plane. This can be seen systematically using a construction known as the magic square, due to Hans Freudenthal and Jacques Tits.

The Tits–Koecher construction produces forms of the E7 Lie algebra from Albert algebras, 27-dimensional exceptional Jordan algebras.

By means of a Chevalley basis for the Lie algebra, one can define E7 as a linear algebraic group over the integers and, consequently, over any commutative ring and in particular over any field: this defines the so-called split (sometimes also known as “untwisted”) adjoint form of E7. Over an algebraically closed field, this and its double cover are the only forms however, over other fields, there are often many other forms, or “twists” of E7, which are classified in the general framework of Galois cohomology (over a perfect field k) by the set H 1 (k, Aut(E7)) which, because the Dynkin diagram of E7 (see below) has no automorphisms, coincides with H 1 (k, E7, ad). [1]

Over the field of real numbers, the real component of the identity of these algebraically twisted forms of E7 coincide with the three real Lie groups mentioned above, but with a subtlety concerning the fundamental group: all adjoint forms of E7 have fundamental group Z/2Z in the sense of algebraic geometry, meaning that they admit exactly one double cover the further non-compact real Lie group forms of E7 are therefore not algebraic and admit no faithful finite-dimensional representations.

Over finite fields, the Lang–Steinberg theorem implies that H 1 (k, E7) = 0, meaning that E7 has no twisted forms: see below.

Dynkin diagram Edit

Root system Edit

Even though the roots span a 7-dimensional space, it is more symmetric and convenient to represent them as vectors lying in a 7-dimensional subspace of an 8-dimensional vector space.

Note that the 7-dimensional subspace is the subspace where the sum of all the eight coordinates is zero. There are 126 roots.

They are listed so that their corresponding nodes in the Dynkin diagram are ordered from left to right (in the diagram depicted above) with the side node last.

An alternative description Edit

An alternative (7-dimensional) description of the root system, which is useful in considering E7 × SU(2) as a subgroup of E8, is the following:

and the two following roots

Thus the generators consist of a 66-dimensional so(12) subalgebra as well as 64 generators that transform as two self-conjugate Weyl spinors of spin(12) of opposite chirality, and their chirality generator, and two other generators of chiralities ± 2 >> .

Given the E7 Cartan matrix (below) and a Dynkin diagram node ordering of:

Weyl group Edit

The Weyl group of E7 is of order 2903040: it is the direct product of the cyclic group of order 2 and the unique simple group of order 1451520 (which can be described as PSp6(2) or PSΩ7(2)). [2]

Cartan matrix Edit

E7 has an SU(8) subalgebra, as is evident by noting that in the 8-dimensional description of the root system, the first group of roots are identical to the roots of SU(8) (with the same Cartan subalgebra as in the E7).

In addition to the 133-dimensional adjoint representation, there is a 56-dimensional "vector" representation, to be found in the E8 adjoint representation.

The characters of finite dimensional representations of the real and complex Lie algebras and Lie groups are all given by the Weyl character formula. The dimensions of the smallest irreducible representations are (sequence A121736 in the OEIS):

1, 56, 133, 912, 1463, 1539, 6480, 7371, 8645, 24320, 27664, 40755, 51072, 86184, 150822, 152152, 238602, 253935, 293930, 320112, 362880, 365750, 573440, 617253, 861840, 885248, 915705, 980343, 2273920, 2282280, 2785552, 3424256, 3635840.

The underlined terms in the sequence above are the dimensions of those irreducible representations possessed by the adjoint form of E7 (equivalently, those whose weights belong to the root lattice of E7), whereas the full sequence gives the dimensions of the irreducible representations of the simply connected form of E7. There exist non-isomorphic irreducible representation of dimensions 1903725824, 16349520330, etc.

The fundamental representations are those with dimensions 133, 8645, 365750, 27664, 1539, 56 and 912 (corresponding to the seven nodes in the Dynkin diagram in the order chosen for the Cartan matrix above, i.e., the nodes are read in the six-node chain first, with the last node being connected to the third).

E7 Polynomial Invariants Edit

E7 is the automorphism group of the following pair of polynomials in 56 non-commutative variables. We divide the variables into two groups of 28, (p, P) and (q, Q) where p and q are real variables and P and Q are 3×3 octonion hermitian matrices. Then the first invariant is the symplectic invariant of Sp(56, R):

The second more complicated invariant is a symmetric quartic polynomial:

C 2 = ( p q + T r [ P ∘ Q ] ) 2 + p T r [ Q ∘ Q

An alternative quartic polynomial invariant constructed by Cartan uses two anti-symmetric 8x8 matrices each with 28 components.

The points over a finite field with q elements of the (split) algebraic group E7 (see above), whether of the adjoint (centerless) or simply connected form (its algebraic universal cover), give a finite Chevalley group. This is closely connected to the group written E7(q), however there is ambiguity in this notation, which can stand for several things:

  • the finite group consisting of the points over Fq of the simply connected form of E7 (for clarity, this can be written E7,sc(q) and is known as the “universal” Chevalley group of type E7 over Fq),
  • (rarely) the finite group consisting of the points over Fq of the adjoint form of E7 (for clarity, this can be written E7,ad(q), and is known as the “adjoint” Chevalley group of type E7 over Fq), or
  • the finite group which is the image of the natural map from the former to the latter: this is what will be denoted by E7(q) in the following, as is most common in texts dealing with finite groups.

From the finite group perspective, the relation between these three groups, which is quite analogous to that between SL(n, q), PGL(n, q) and PSL(n, q), can be summarized as follows: E7(q) is simple for any q, E7,sc(q) is its Schur cover, and the E7,ad(q) lies in its automorphism group furthermore, when q is a power of 2, all three coincide, and otherwise (when q is odd), the Schur multiplier of E7(q) is 2 and E7(q) is of index 2 in E7,ad(q), which explains why E7,sc(q) and E7,ad(q) are often written as 2·E7(q) and E7(q)·2. From the algebraic group perspective, it is less common for E7(q) to refer to the finite simple group, because the latter is not in a natural way the set of points of an algebraic group over Fq unlike E7,sc(q) and E7,ad(q).

As mentioned above, E7(q) is simple for any q, [3] [4] and it constitutes one of the infinite families addressed by the classification of finite simple groups. Its number of elements is given by the formula (sequence A008870 in the OEIS):

The order of E7,sc(q) or E7,ad(q) (both are equal) can be obtained by removing the dividing factor gcd(2, q−1) (sequence A008869 in the OEIS). The Schur multiplier of E7(q) is gcd(2, q−1), and its outer automorphism group is the product of the diagonal automorphism group Z/gcd(2, q−1)Z (given by the action of E7,ad(q)) and the group of field automorphisms (i.e., cyclic of order f if q = p f where p is prime).

N = 8 supergravity in four dimensions, which is a dimensional reduction from 11 dimensional supergravity, admit an E7 bosonic global symmetry and an SU(8) bosonic local symmetry. The fermions are in representations of SU(8), the gauge fields are in a representation of E7, and the scalars are in a representation of both (Gravitons are singlets with respect to both). Physical states are in representations of the coset E7 / SU(8) .

In string theory, E7 appears as a part of the gauge group of one of the (unstable and non-supersymmetric) versions of the heterotic string. It can also appear in the unbroken gauge group E8 × E7 in six-dimensional compactifications of heterotic string theory, for instance on the four-dimensional surface K3.


Mathematics Solutions for Class 7 Math Chapter 6 - Indices

Mathematics Solutions Solutions for Class 7 Math Chapter 6 Indices are provided here with simple step-by-step explanations. These solutions for Indices are extremely popular among Class 7 students for Math Indices Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Solutions Book of Class 7 Math Chapter 6 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Solutions Solutions. All Mathematics Solutions Solutions for class Class 7 Math are prepared by experts and are 100% accurate.

Page No 44:

Question 1:

Sr. No. Indices
(Numbers in index
form)
Base Index Multiplication form Value
(i) 3 4 3 4 3 × 3 × 3 × 3 81
(ii) 16 3
(iii) ( - 8) 2
(iv) 3 7 × 3 7 × 3 7 × 3 7 81 2401
(v) ( - 13) 4

Answer:

Sr. No. (Numbers in index form) Base Index Multiplication form Value
(i) 3 4 3 4 3 × 3 × 3 × 3 81
(ii) 16 3 16 3 16 × 16 × 16 4096
(iii) (&minus8) 2 (&minus8) 2 (&minus8) × (&minus8) 64
(iv) 3 7 4 3 7 4 3 7 × 3 7 × 3 7 × 3 7 81 2401
(v) (&minus13) 4 &minus13 4 (&minus13) × (&minus13) × (&minus13) × (&minus13) 28561

Page No 44:

Question 2:

Answer:


i   2 10 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024
ii   5 3 = 5 × 5 × 5 = 125
iii   - 7 4   = - 7 × - 7 × - 7 × - 7 = 2401
iv   - 6 3 = - 6 × - 6 × - 6 = - 216
v   9 3 = 9 × 9 × 9 =   729
vi   8 1 = 8 = 8
vii   4 5 3 = 4 5 × 4 5 × 4 5 = 64 125
viii   - 1 2 4 = - 1 2 × - 1 2 × - 1 2 × - 1 2 = 1 16

Page No 45:

Question 1:

(i) 7 4 × 7 2 (ii) ( - 11) 5 × ( - 11) 2 (iii) 6 7 3   ×   6 7 5  


(iv) - 3 2 5   ×   - 3 2 3   (v) a 16   ×   a 7 (vi) P 5 3   × P 5 7

Answer:

It is known that, a m × a n = a m+n , where m and n are integers and a is a non-zero rational number.
i   7 4 × 7 2 =   7 4 + 2 = 7 6
ii   - 11 5 × - 11 2 = - 11 5 + 2 = - 11 7
iii   6 7 3 × 6 7 5 = 6 7 3 + 5 = 6 7 8
iv   - 3 2 5 × - 3 2 3 = - 3 2 5 + 3 = - 3 2 8
v   a 16 × a 7 = a 16 + 7 = a 23
vi   p 5 3 × p 5 7 = p 5 3 + 7 = p 5 10

Page No 46:

Question 1:

Answer:

It is known that, a m ÷ a n = a m&minusn , where m and n are integers and a is non-zero rational number.
i   a 6 ÷ a 4 = a 6 - 4 = a 2
ii   m 5 ÷ m 8 = m 5 - 8 = m - 3
iii   p 3 ÷ p 13 = p 3 - 13 = p - 10
iv   x 10 ÷ x 10 = x 10 - 10 = x 0 = 1         ∵   a 0 = 1

Page No 46:

Question 2:

(i) ( - 7 ) 12   ÷ ( - 7 ) 12 (ii) 7 5 ÷   7 3 (iii) 4 5 3 ÷ 4 5 2 (iv) 4 7 ÷ 4 5

Answer:

It is known that, a m ÷ a n = a m&minusn , where m and n are integers and a is a non-zero rational number.
i   - 7 12 ÷ - 7 12 = - 7 12 - 12 = - 7 0 = 1         ∵   a 0 = 1
ii   7 5 ÷ 7 3 = 7 5 - 3 = 7 2 = 7 × 7 = 49
iii   4 5 3 ÷ 4 5 2 = 4 5 3 - 2 = 4 5 1 = 4 5
iv   4 7 ÷ 4 5 = 4 7 - 5 = 4 2 = 4 × 4 = 16

Page No 48:

Question 1:

(i) 15 12 3 4 (ii) 3 4 - 2 (iii) 1 7 - 3 4 (iv) 2 5 - 2 - 3 (v) 6 5 4

(vi) 6 7 5 2 (vii) 2 3 - 4 5 (viii) 5 8 3 - 2 (ix) 3 4 6 1 (x) 2 5 - 3 2

Answer:

It is known that, (a m ) n = a mn , where m and n are integers and a is a non-zero rational number.
i   15 12 3 4 = 15 12 3 × 4 = 15 12 12
ii   3 4 - 2 = 3 4 × - 2 = 3 - 8
iii   1 7 - 3 4 = 1 7 - 3 × 4 = 1 7 - 12
iv   2 5 - 2 - 3 = 2 5 - 2 × - 3 = 2 5 6
v   6 5 4 = 6 5 × 4 = 6 20
vi   6 7 5 2 = 6 7 5 × 2 = 6 7 10
vii   2 3 - 4 5 = 2 3 - 4 × 5 = 2 3 - 20
viii   5 8 3 - 2 = 5 8 3 × - 2 = 5 8 - 6
ix   3 4 6 1 = 3 4 6 × 1 = 3 4 6
x   2 5 - 3 2 = 2 5 - 3 × 2 = 2 5 - 6

Page No 48:

Question 2:

Write the following numbers using positive indices.

(i) 2 7 - 2 (ii) 11 3 - 5 (iii) 1 6 - 3 (iv) y - 4

Answer:

It is known that, a - m = 1 a m where m is an integer and a is a non-zero rational number.
i   2 7 - 2 = 1 2 7 2 = 7 2 2
ii   11 3 - 5 = 1 11 3 5 = 3 11 5
iii   1 6 - 3 = 1 1 6 3 = 6 1 3 = 6 3
iv   y - 4 = 1 y 4 = 1 y 4

Page No 50:

Question 1:

Find the square root.
(i) 625 (ii) 1225 (iii) 289 (iv) 4096 (v) 1089

Answer:


(i) The prime factorization of 625 is,
625 = 5 × 5 × 5 × 5
To find the square root, we will take one number from each pair and multiply.
625 = 5 × 5 = 25 ∴ 625 = 25
(ii) The prime factorization of 1225 is,
1225 = 5 × 5 × 7 × 7
To find the square root, we will take one number from each pair and multiply.
1225 = 5 × 7 = 35 ∴ 1225 = 35
(iii) The prime factorisation of 289 is,
289 = 17 × 17
To find the square root, we will take one number from each pair and multiply.
289 = 17 × 17 = 17 ∴ 289 = 17
(iv) The prime factorization of 4096 is,
4096 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 ​× 2
To find the square root, we will take one number from each pair and multiply.
4096 = 2 × 2 × 2 × 2 × 2 × 2   =   64 ∴ 4096 = 64
(v) The prime factorizationn of 1089 is,
1089 = 3 × 3 × 11 × 11
To find the square root, we will take one number from each pair and multipy.
1089 = 3 × 11 = 33 ∴ 1089 = 33


EBook Do the Math 7-5 Comparing APRs James Sprater of Grand Junction, Colorado, has been shopping for a loan to buy a used car. He wants to borrow $18,000 for four or five years. James' credit union offers a declining-balance loan at 9.3 percent for 48 months, resulting in a monthly payment of $450.50. The credit union does not offer five-year auto loans for amounts less than $20,000, however. If James borrowed $18,000, this payment would strain his budget. A local bank offered current depositors a five-year loan at a 9.34 percent APR, with a monthly payment of $376.62. This credit would not be a declining-balance loan. Because James is not a depositor in the bank, he would also be charged a $25 credit check fee and a $60 application fee. James likes the lower payment but knows that the APR is the true cost of credit, so he decided to confirm the APRs for both loans before making his decision. Round your answers to two decimal places. What is the APR for the credit union loan? % Use the n-ratio formula to calculate and confirm the APR on the bank loan as quoted for depositors. % What is the add-on interest rate for the bank loan? % What would be the true APR on the bank loan if James did not open an account to avoid the credit check and application fees? %

James Sprater of Grand Junction, Colorado, has been shopping for a loan to buy a used car. He wants to borrow $18,000 for four or five years. James' credit union offers a declining-balance loan at 9.3 percent for 48 months, resulting in a monthly payment of $450.50. The credit union does not offer five-year auto loans for amounts less than $20,000, however. If James borrowed $18,000, this payment would strain his budget. A local bank offered current depositors a five-year loan at a 9.34 percent APR, with a monthly payment of $376.62. This credit would not be a declining-balance loan. Because James is not a depositor in the bank, he would also be charged a $25 credit check fee and a $60 application fee. James likes the lower payment but knows that the APR is the true cost of credit, so he decided to confirm the APRs for both loans before making his decision. Round your answers to two decimal places.

What is the APR for the credit union loan?

Use the n-ratio formula to calculate and confirm the APR on the bank loan as quoted for depositors.

What is the add-on interest rate for the bank loan?

What would be the true APR on the bank loan if James did not open an account to avoid the credit check and application fees?


5-Year Mortgage Calculator

5-Year Mortgage Calculator is an online personal finance assessment tool to calculate monthly repayment, total repayment and total interest cost on the principal borrowed. The loan amount and interest rate are the key terms of 5-year mortgage to calculate the necessary repayment details.

5-year mortgage is a home loan that will enable you to purchase a house and expect to have repaid in five years. Five year mortgages can be the solution for some people wanting to buy a home. Before buying or refinancing with a five year mortgage it is very important to do a research to calculate how the future loan payments are involved and how these payments may affect your financial situation. With interest rates rising and real estate prices booming, lenders are starting to offer the 5-year mortgage as a viable option for buying your dream home. There is various money lenders are participating in real estate business nowadays with different terms of conditions. Therefore, analyzing the 5 mortgage components is a vital step before opt for any home loan agreement. When it comes to online calculation 5-year mortgage calculator assists you to determine which lender provides you best option by comparing different interest rates and loan amounts available in finance market.


Amortization Calculation

Usually, whether you can afford a loan depends on whether you can afford the periodic payment (commonly a monthly payment period). So, the most important amortization formula is the calculation of the payment amount per period.

Calculating the Payment Amount per Period

The formula for calculating the payment amount is shown below.

  • A = payment Amount per period
  • P = initial Principal (loan amount)
  • r = interest rate per period
  • n = total number of payments or periods

Example: What would the monthly payment be on a 5-year, $20,000 car loan with a nominal 7.5% annual interest rate? We'll assume that the original price was $21,000 and that you've made a $1,000 down payment.

You can use the amortization calculator below to determine that the Payment Amount (A) is $400.76 per month.

P = $20,000
r = 7.5% per year / 12 months = 0.625% per period (this is entered as 0.00625 in the calculator)
n = 5 years * 12 months = 60 total periods

Amortization Payment Calculator

Calculating the Monthly Payment in Excel

Microsoft Excel has a number of built-in functions for amortization formulas. The function corresponding to the formula above is the PMT function. In Excel, you could calculate the monthly payment using the following formula:

Calculating the Rate Per Period

When the number of compounding periods matches the number of payment periods, the rate per period (r) is easy to calculate. Like the above example, it is just the nominal annual rate divided by the periods per year. However, what do you do if you have a Canadian mortage and the compounding period is semi-annual, but you are making monthly payments? In that case, you can use the following formula, derived from the compound interest formula.

  • r = rate per payment period
  • i = nominal annual interest rate
  • n = number of compounding periods per year
  • p = number of payment periods per year

Example: If the nominal annual interest rate is i = 7.5%, and the interest is compounded semi-annually ( n = 2 ), and payments are made monthly ( p = 12 ), then the rate per period will be r = 0.6155%.

Important: If the compound period is shorter than the payment period, using this formula results in negative amortization (paying interest on interest). See my article, "negative amortization" for more information.

If you are trying to solve for the annual interest rate, a little algebra gives:

Example: Using the RATE() formula in Excel, the rate per period (r) for a Canadian mortgage (compounded semi-annually) of $100,000 with a monthly payment of $584.45 amortized over 25 years is 0.41647% calculated using r=RATE(25*12,-584.45,100000) . The annual rate is calculated to be 5.05% using the formula i=2*((0.0041647+1)^(12/2)-1) .


Understanding An Amortization Schedule

By committing to a mortgage loan, the borrower is entering into a financial agreement with a lender to pay back the mortgage money, with interest, over a set period of time.

The borrower’s monthly mortgage payment may change over time depending on the type of loan program, however, we’re going to address the typical 30 year fixed Principal and Interest loan program for the sake of breaking down the individual payment components for this particular article about an amortization schedule.

On each payment that is made, a certain amount of interest is taken out to pay the lender back for the opportunity to borrow the money, and the remaining balance is applied to the principal balance.

It’s common to hear industry professionals and homeowners talk about a mortgage payment being front-loaded with interest, especially if they’re referencing an amortization chart to show the numbers. Since there is more interest being paid at the beginning of a mortgage payment term the amount of money applied to interest decreases over time, while the money applied to the principal increases.

The Loan Amortization Chart

We can better understand mortgage payments by looking at a loan amortization chart, which shows the specific payments associated with a loan.

The details will include the interest and principal component of each periodic payment.

For example, let’s look at a scenario where you borrowed a $100,000 loan at 7.5% interest rate, fixed for 30 year term. To ensure full repayment of principal by the end of the 30 years, your payment would need to be $699.21 per month. In the first month, you owe $100,000, which means the interest would be calculated on the full loan amount. To calculate this, we start with $100,000 and multiply it by 7.5% interest rate. This will give you $7,500 of annual interest. However, we only need a monthly amount. So we divide by 12 months to find that the interest equals $625. Now remember, you are paying $699.21. If you only owe interest of $625, then the remainder of the payment, $74.21, will go towards the principal. Thus, your new outstanding balance is now $99,925.79.

In month #2, you make the same payment of $699.21. However, this time, you now owe $99,925.79. Therefore, you will only pay interest on $99,925.79. When running through the calculator in the same process detailed above, you will find that your interest component is $624.54. (It is decreasing!) The remaining $74.68 will be applied towards principal. (This amount is increasing!)

Each month, the same simple mathematic calculation will be made. Because the payments are remaining the same, each month the interest will continue to be reduced and the remainder going towards principal will continue to increase.

An amortization chart runs chronologically through your series of payments until you get to the final payment. The chart can also be a useful tool to determine interest paid to date, principal paid to date, or remaining principal.

Another frequent use of amortization charts is to determine how extra payments toward principal can affect and accelerate the month of final payment of the loan, as well as reduce your total interest payments.


How to Get the Best Loan Interest Rates

One of the easiest ways to make sure you’re getting the best interest rate is to shop around. Compare loan offers side-by-side, and choose the one that works best for you.

But getting the best personal loan rates actually starts long before you go to take out a loan. It’s the hard work of improving your financial health and credit score before you need to borrow more money.

“The biggest one is to bring down your existing debt,” Nayar says. “The cheapest way to get money is to have money.”

That’s because lenders see less risk in borrowers with less debt, and are willing to offer lower interest rates because of it.

Another option is to bring on a co-signer, someone with better credit who can vouch for you on the loan application. Keep in mind, however, that the co-signer is equally liable for the debt, and it could sap their credit score if you miss payments.

Don’t let the very idea of calculating loan interest and diving into algebraic formulas scare you. Understanding how interest works is a crucial step to making smart decisions about loans.

So whether you break out a pencil and paper, or use one of NextAdvisor’s online calculators, take the time to understand the real cost — interest included — behind your next loan.