|
|
|
|
Home > Word Problems calculators
|
|
|
Educational Level
|
Secondary school, High school and College |
|
Program Purpose
|
Provide step by step solutions of your problems using online calculators (online solvers)
|
|
Problem Source
|
Your textbook, etc |
|
1.
Arithmetic Progression
|
1. For given arithemetic progression series 1,4,7,10,13 ,... find 10 th term and addition of first 10 th terms.
Here first term a = 1,
d = 4 - 1 = 3
We know that, f(n) = a + (n - 1)d
f(10) = 1 + (10 - 1)(3)
= 1 + (27)
= 28
We know that, S_n = n/2 [2a + (n - 1)d]
∴ S_10 = 10/2 [2(1) + (10 - 1)(3)]
= 5 [2 + (27)]
= 5 [29]
= 145
Hence, 10th Term Of The Given Series is 28 And Sum of First 10th Term is 145
2. For arithemetic progression addition of 3 terms is 27 and their multiplication is 648, then that nos.
3. Find the sum of all natural nos between 100 to 200 and which are not divisible by 4.
4. For arithemetic progression, addition of three terms is 51 and multiplication of end terms is 273, then find that nos.
5. For arithemetic progression, addition of 4 terms is 4 and addition of multiplication of end terms and multiplication of middle terms is -38, then find that nos.
|
|
|
2.
Geometric Progression
|
1. For given geometric progression series 3,6,12,24,48,... find 10 th term and addition of first 10 th terms.
Here a = 3,
r = 6 / 3 = 2
We know that, a_n = a * r^(n-1)
a_10 = 3 * (2)^(10 - 1)
= 3 * (512)
= 1536
We know that, S_n = a (r^n - 1)/(r - 1)
∴ S_10 = 3 * ((2)^10 - 1) / (2 - 1)
=> S_10 = 3 * (1024 - 1) / 1
=> S_10 = 3 * 1023 / 1
=> S_10 = 3069
Hence, 10th term of the given series is 1536 and sum of first 10th term is 3069
2. For geometric progression addition of 3 terms is 26 and their multiplication is 216, then that nos.
3. For geometric progression multiplication of 5 terms is 1 and 5th term is 81 times then the 1st term.
4. Arithmetic mean of two no is 13 and geometric mean is 12, then find that nos.
5. Find 6 geometric mean between 1 and 256.
6. Find 12 + 22 + 32 + ... + 102
7. Find 13 + 23 + 33 + ... 10 terms
|
|
|
|
|
3.
Simple Interest
|
1. Find the simple intereset on Rs. 730 for 184 days at 25/4 % per annum.
Here P = Rs. 730 , R = 25/4 % and Time = 184 days = 184 / 365 years
S.I. = P*R*N/100 = ( 730 * 25/4 * 184 /365) / 100 = 23
Simple Interest is Rs. 23 .
2. The interest on a certain amount of money at 8% per year for a period of 4 years is Rs 512. Find the sum of money.
3. A sum of money lent at simple interest amounts to Rs 1596 in 3/2 years and to 1860 in 5/2 years. Find the sum & the rate of interest.
4. A sum was put at simple interest at a certain rate for 2 years. Had it been put at 3 % highere rate, it whould have fetched Rs. 300 more. Find the sum.
5. A shopkeeper borrowed Rs. 20000 from two money lenders. For one loan he paid 12% and for the other 14% per annum. After one year, he paid Rs. 2560 as interest. How much did he borrow at each rate ?
6. At what rate percent per annum will sum of money double in 8 years?
7. Rajeev deposited money in the post office which is doubled in 20 years at a simple rate of interest. In how many years will the original sum triple itself?
|
|
|
4.
Compound Interest
|
1. Calculate Compound Interest and amount on Rs 4500 at 10 % per annum in 3 years.
Here P = Rs. 4500 , R = 10 %, N = 3 yrs.
A = P (1 + R/100)^N
= 4500 * ( 1 + 10 / 100 ) ^ 3
= 4500 * ( 110 / 100 ) ^ 3
= 5989.5
Therefore, C.I. = Rs. ( 5989.5 - 4500 ) = Rs. 1489.5 .
2. What sum of money becomes Rs. 9261 in 3 years at 5% per annum, compounded annually ?
3. A sum of money amounts to Rs 6690 after 3 years and to Rs 10035 after 6 years on compound interest. Calculate the sum of money.
4. The difference between the compound interest and the simple interest on a certain sum at 10 % per annum for 2 years is Rs. 52 . Find the sum.
5. If the compound interest on a certain sum for 3 years at 10 % per annum be Rs. 331, what would be the simple interest ?
6. A sum of money 2 times itself at compound interest in 15 years. In how many years, it will become 8 times of itself ?
|
|
|
|
|
5.
HCF-LCM Word Problem
|
1. The HCF of two nos is 14 and their LCM is 11592 . If one of the nos is 504 , find the other?
Other no = ( HCF * LCM ) / Given No = ( 14 * 11592 ) / 504 = 322 .
2. Find the largest no which can exactly divide 513 , 783 , 1107
3. Find the smallest no exactly divisible by 12 , 15 , 20 and 27 .
4. Find the least no which when divided by 6 , 7 , 8 , 9 , 12 leaves the same remainder 2 in each case.
5. Find the largest no which divides 77 , 147 , 252 to leave the same remainder in each case.
6. The greatest no that will divide 290 , 460 , 552 leaving respectively 4 , 5 , 6 as remainder.
7. LCM of two nos is 14 times their HCF. The sum of LCM and HCF is 600 . If one no is 280 , then find the other no ?
|
|
|
6.
Installment
|
1. A briefcase is available for Rs 440 cash or for Rs 200 cash down payment and Rs 244 to be paid after 1 months. Find the rate of intereset charged under the installment plan.
Cash Price = 440
Down Payemnt = 200
Remaining Balance = 440 - 200 = 240
The installment to be paid at the end of 1 months = 244
Therefore the interest charged on Rs 240, for a period of 1 months = Rs 244 - Rs 240 = 4
If R % is the rate of interest per annum, then
(240 × R × 1) / (100 × 12) = 4
R = 20
Thus, the rate of interest charged under the installment plan is = 20 per annum
2. A washing machine is available at Rs 6400 cash or for Rs 1400 cash down payment and 5 monthly installments of Rs 1030 each. Calculate the rate of interest charged under the instalment plan.
3. A computer is sold by a company for Rs 19200 cash or for Rs 4800 cash down payment together with 5 equal monthly installments. If the rate of interest charged by the company is 12% per annum, find each installment.
4. A man borrows money from a finance company and has to pay it back in 2 equal half yearly installments of Rs 4945 each. If the interest is charged by the finance company at the rate of 15 % per annum compounded as installment plan, find the principal and the total interest paid.
5. Ram borrowed a sum of money and returned it in 3 equal quarterly installments of Rs 17576 each. Find the sum borrowed, if the rate of interest charged was 16 % per annum compounded as installment plan. Find also the total interest charged.
|
|
|
|
|
7.
Percentage
|
1. A number is 20 % of 80 .
20 % of 80 = ( 20 / 100 × 80 ) = 16 .
2. A's annual income is increased from Rs 60000 to Rs 75000 . Find the percentage of increase in A's income.
3. In a school of 225 boys, 15 were absent then what percent were present ?
4. A earns 25 % more than B. By what percent does B earn less then A.
5. A reduction of 20 % in the price of basmati rice would enable a man to buy 2 kg of rice more for Rs 250. Find the reduced price per kg.
6. Find the selling price of an item, of which the printed price is Rs 25000 if the successive discounts given are 10 %, 8 % and 4 %.
7. The successive discount of 10 % and 5 % are given on the purchased Computer. If the final price of the Computer is Rs 10260, then find the printed price of the Computer.
|
|
|
8.
Profit Loss Discount
|
1.Find selling price when cost price = Rs. 50 , Loss/Gain = 10 % .
C.P. = Rs. 50 , Gain = 10 %
S.P. = C.P. × ( 100 + Gain% ) / 100 = 50 × ( 100 + 10 ) / 100 = 55 Rs.
C.P. = Rs. 50 , Loss = 10 %
S.P. = C.P. × ( 100 - Loss% ) / 100 = 50 × ( 100 - 10 ) / 100 = 45 Rs.
2. Find cost price when selling price = 55 , Loss/Gain = 10 %
3. Find Loss/Gain % when cost price = 50 and selling price = 55
4. By selling an article for Rs 110 a man loses 12 %. For how much should he sell it to gain 8 %.
5. A man sells two houses for Rs 536850 each. On one he gains and on the other he loses 5 %. Find his gain or loss % on the whole transaction.
6. If selling price of 10 articles is the same as the cost price of 12 articles, find the gain %.
7. If the marked price of an article is Rs. 380 and a discount of 5 % is given on it, what is the selling price ?
8. A cycle dealer marks his goods 25 % above his cost price and allows a discount of 8 % on it. Find his gain percent.
9. Successive discounts of 20 % and 10 % equivalent to a single discount of how many percent?
|
|
|
|
|
9.
Statistics Word Problem
|
1. The mean of 10 observations is 12.5 . While calculating the mean
one observation was by mistake taken as (-8) instead of (+8) . Find the correct
mean.
Here Mean X = 12.5
and n=10
`Sigma`(X) = X × n = 12.5 × 10 = 125
`:.` the correct sum `Sigma`(X) = 125 - (Wrong Observation) + (Correct Observation)
= 125 - (-8) + (8) = 141 .
`:.` correct mean = correct
sum / n = 141 /
10 = 14.1
2. The sum of 15 observation is 343 . If we remove two
observation 18 and 26 , then find out the mean of remaining
observations.
Here `Sigma`(X) = 343 and n= 15
If we remove two observation then 13 observation left.
Sum of 13 observation `Sigma`(X') = 343 - ( ( 18 ) + ( 26 ) ) = 299
Then the mean of remaining 13 observations = `Sigma`(X') / n = 299 / 13 = 23.
and many more...
|
|
|
|
|
|
|
|
|
|
|
|
|
Share this solution or page with your friends.
|
|
|
|
