Question 1

683.46 - 227.39 - 341.85 =?

Accepted Answer

? = 683.46 - 227.39 - 341.85 = 683.46 - 569.24 = 114.22

Question 2

Mohan gets 3 marks for each correct sum and loses 2 marks for each wrong sum. He attempts 30 sums and obtains 40 marks. The number of sums solved correctly is

Accepted Answer

Solution 1. Let the question solved correctly by Mohan = X ; and number of question solved wrongly by Mohan = 30 - X ; According to the question, 3X - (30 - X) x 2 = 40 ⇒ 3x - 60 + 2x = 40 ⇒ 5x - 60 = 40 ⇒ 5x = 40 + 60 ⇒ x = 100/5 = 20 ∴ Mohan attempted 20 questions correctly . Solution 2. Let the question solved correctly by Mohan = X ; and number of question solved wrongly by Mohan = Y ; According to given question; X + Y = 30 ; -----------i 3X - 2Y = 40 ;-------ii Now multiply the first equation by 2; we will get 2X + 2Y = 60 ;--------iii After Add the equation ii and iii , we will get 3X - 2Y + 2X + 2Y = 40 + 60 ; ⇒ 5X = 100 ; ⇒ X = 20; ∴ Mohan attempted 20 questions correctly .

Question 3

It is required to get 40% marks to qualify an exam. A candidate scored 200 marks and failed by 8 marks. What were the maximum marks of that exam?

Accepted Answer

Let the maximum marks = x. According to the question, 40x/100 = 200 + 8 ⇒ x = 208 x 10/4 = 520

Question 4

One-third of Rahul's marks in Mathematics exceeds one-half of his marks in Hindi by 30. If he got 480 marks in the two subjects together, how many marks did he get in Hindi ?

Accepted Answer

Let Rahul's marks in Mathematics = x and in Hindi = y According to the question, 1/3 x - 1/2y = 30 ⇒ 2x - 3y = 180 ...(i) Also, given x + y = 480 ...(ii) By solving Eqs. (i) and (ii), we get x = 324 and y =156

Question 5

A container of milk was 4/5 full. When 12 bottles of milk were taken out and 8 bottles of milk were poured into it, it was 3/4 full. How many bottles of milk can the container contain?

Accepted Answer

Let x bottles can fill the container completely. According to the questions, 4/5 x - 3/4 x = (12 - 8) ⇒ 16x - 15x/20 = (12 - 8) ⇒ x/20 = 4 ∴ x = 80 ∴ Required number of bottles = 80.

Question 6

The number obtained by interchanging the digits of a two-digit number is more than the original number by 27 and the sum of the digits is 13. What is the original number ?

Accepted Answer

Let the ten's place digit = x and unit's place digit = y ∴ The original number = 10x + y After interchanging the digits Number will become 10y + x ∴ according to the given question, y + x = 13 and (10y + x ) - (10x + y) = 27 ⇒ 9y - 9x = 27 ⇒ y - x = 3 ................(i) and y + x = 13 .................(ii) On solving Eqs.(i) and (ii) , we get y = 8 and x = 5 ∴ Required number is 10x + y = 10 x 5 + 8 = 58

Question 7

Of three positive numbers the product of the first and second is 42, that of the second and third is 56 and that of third and first is 48, The third number is

Accepted Answer

Let first, second and third numbers be x , y and z, respectively. Then , xy = 42.................(i) yz = 56 ..................(ii) xz = 48 ....................(iii) Multiplying Eqs ( i), (ii) and (iii), we get (xyz)2 = 42 x 56 x 48 ⇒ (xyz)2 = 112896 ⇒ xyz = 336 ................(iv) Dividing Eq. (iv) by Eq. (i), we get xyz/xy = 336/42 ⇒ z = 8

Question 8

A man spends 2/5 of his income on house rent, 3/10 on food and 1/8 on conveyance. If he has ₹ 2800 left with him, find his expenditure on food and conveyance together.

Accepted Answer

Hint 1. Let consider the monthly income = x ; Amount spend on House rent = 2x/5; Amount spend on food = 3x/10; Amount spend on conveyance = x/8; After Expenditure , The left amount = 2800; According to Question; x - ( 2x/5 + 3x/10 + x/8 ) = 2800 ; x- ((16x + 12x + 5x)/40) = 2800 ; x - 33x/40 = 2800 ; (40x - 33x)40 = 2800 ; 7x/40 = 2800 ; 7x = 2800 x 40 ; x = 2800 x 40/7 ; x = 400 x 40 ; x = 16000 ; Expenditure on food = 3x/10 = 3 x 16000 /10= ? 4800 Expenditure on conveyance = x/8 = 1 x 16000/8 = ? 2000 ? Expenditure on food and conveyance together =? (4800 + 2000) = ? 6800 Solution 2 . part of income left = 1 - (2/5 + 3/10 + 1/8 ) =1- (16 + 12 + 5/40) = 1- 33/40 = 40 - 33/40 = 7 /40 Let the monthly income = x According to the question, 7/40 x = 2800 ∴ x = 2800 x 40/7 = 400 x 40 = ₹ 16000 Expenditure on food = 3/10 x 16000 = ₹ 4800 Expenditure on conveyance = 1/8 x 16000 = ₹ 2000 ∴ Expenditure on food and conveyance together =₹ (4800 + 2000) = ₹ 6800

Question 9

Mani Ram divides ₹ 17200 among his 5 sons, 4 daughters and 2 friends. If each daughter receives four times as much as each friend receives and each son receives five times as mush as each friend receives, how much does each daughter receive?

Accepted Answer

Let share of each friend = a. Then, share of each daughter = 4a. and share of each son = 5a. According to the question, 5 x 5a + 4 x 4a + 2 x a = 17200 ⇒ 25x + 16x + 2x = 17200 ⇒ 43x = 17200 ∴ x = 17200/43 = 400 ∴ Share of each daughter = 4x = 4 x 400 = ₹ 1600

Question 10

From a group of men and women, 15 women leave from it, then there are 2 men for each women. After this, 45 men leave , then there are 5 women for each men. Find the double of the number of women in the beginning.?

Accepted Answer

Let at present, there are c men, then number of women = 5c Before the men had left number of men = c + 45 and number of women = 5c ∴ c + 45 = 2 x 5c ⇒ 9c = 45 ∴ c = 45/9 = 5 Hence, number of women in the beginning = 5c + 15 = 5 x 5 + 15 = 25 + 15 = 40 ∴ Double of the women = 40 x 2 = 80

Question 11

Bus fare between Raipur and Mirpur for one adult is six times the fare for one child. If an adult's bus fare is ₹ 114, how much amount will be paid by 4 adults and 5 children together for travelling the same distance?

Accepted Answer

Adult fare = ₹ 114 According to the question, child fare = Adult fare x 1/6 = 144 x 1/6 = ₹ 19 ∴ Total amount paid by 4 adults and 5 children are for the same distance. = ₹(4 x 144 + 5 x 19) = ₹(456 + 95) = ₹ 551

Question 12

The total cost of 12 pens and 5 pencils is equal to ₹ 111. Also, the cost of 1 pencil is ₹ 5 less than the cost of 1 pen. What will be the cost of 8 pens and 9 pencils?

Accepted Answer

Let the cost of 1 pencil = x ∴ Cost of 1 pen = (x + 5) According to the question, 12(x + 5) + 5x = 111 ⇒ 12x + 60 + 5x = 111 ⇒ 17x = 111 - 60 = 51 ⇒ x = 51/17 = 3 ∴ Cost of 1 pencil = ₹ 3 Cost of 1 pen = ₹ 8 ∴ Total Cost of 8 pens and 9 pencils = (8 x 8 + 9 x 3) = ₹(64 + 27) = ₹ 91

Question 13

When an amount is distributed amongst 14 girls, each of them gets ₹ 160 more than the amount received by each girl in that condition when the same amount t is distributed equally amongst 18 girls. Find the amount.

Accepted Answer

Let the total amount = x According to the question. 14 girls are getting x amount. 1 girl will get x/14 amount. Similarly if 18 girls are getting x amount. 1 girl will get x/18 amount. according to question, x/14 - x/18 = 160 ⇒ (9x - 7x)/126 = 160 ⇒ 2x/126 = 160 ⇒ x/126 = 80 ∴ x = 126 x 80 = ₹ 10080

Question 14

A man had 170 currency notes in all, some of which were of ₹ 100 denominations and some of ₹ 50 denominations. The total amount of all these currency notes was ₹ 10000. How much amount did he have in the denominations of ₹ 50?

Accepted Answer

Solution 1 : Let the number of ₹ 50 notes = x, then the number of ₹ 100 notes = (170 - x) According to question. 50x + 100(170 - x) = 10000 divide by 50 the above equation, we will get ⇒ x + 2(170 - x) = 200 ⇒ x + 340 - 2x = 200 ⇒ -x = 200 - 340 = -140 ⇒ x = 140 ∴ Required amount ₹ 50 x 140 = ₹ 7000 Solution 2 : Let the number of ₹ 50 notes = x and the number of ₹ 100 notes = y. according to given question, x + y = 170 .............(1) 50x + 100y = 10000............(2) Divide the equation (2) by 10 , we will get 5x + 17y = 1000............(3) Multiply the equation (1) by 5 5x + 5y = 850 ------------(4) Subtract the (4) equation from (3) equation, we will get, 5x +10y - ( 5x + 5y ) = 1000 - 850 ⇒ 5x +10y - 5x - 5y = 1000 - 850 ⇒ 10y - 5y = 1000 - 850 ⇒ 5y = 150 ⇒ y = 150/5 ⇒ y = 30 Number of ₹ 100 notes = y = 30; since x + Y = 170 therefore x = 170 - y ⇒ x = 170 - 30 ⇒ x = 140 Number of ₹ 50 notes = x = 140

Question 15

Mr. Sahni employed a worker for a certain work to be done in some days. He pays ₹20 to the worker for each working day and the payment is reduced by ₹ 3 for each non-working day of the worker. For how many days the worker does remain absent from wor…

Accepted Answer

Let the worker remains absent for x days. Then, his working days = (120 - x) days. According to the question, 20 x Working day - 3 x Absent day = 560 20(120 - x) - 3x = 560 ⇒ 2400 - 20x - 3x = 560 ⇒ x = 2400 - 560/23 = 1840/23 = 80 days

Question 16

A fruit seller has a crate of apple containing 1 bruised apple for every 40 apples in the crate, If 3 out of every 4 bruised apples are considered unsaleable and there are 9 unsaleable apples in the crate, then how many apples are there in the crate?

Accepted Answer

Let total number of apples in the crate = a Then, number of bruised apples = a/40 . Number of unsaleable apples since 4 bruise apples containing 3 unsaleable apples. therefore 1 bruise apples will contain 3/4 unsaleable apples hence a/40 bruise apples will contain 3/4 x a/40 unsaleable apples = 3/4 x a/40 = 3a/160 According to the question. 3a/160 = 9 ⇒ a = 9 x 160/3 a = 3 x 160 = 480

Question 17

City A has a population of 136000 which is decreasing at the rate of 2400 per year , City B has a population of 84000 which is increasing at the rate of 1600 per year . In how many years will the population of two cities become equal ?

Accepted Answer

Let population of cities A and B become equal after x year. According to the question. Population of city A = Population of city B Then, 136000 - 2400 x = 84000 + 1600 x. ⇒ 4000x = 52000 ∴ x = 52/4 = 13 year

Question 18

5 - [3/4 +{2 1/ 2 - (1/2 + 1/6 - 1/7 )}]/2 =?

Accepted Answer

? = 5 - [3/4 + {5/2 - (1/2 + 1/6 - 1/7)}]/2 = 5 - [3/4 + {5/2 - (1/2 + 7 - 6/42 )}]/2 = 5 - [3/4 + {5/2 - (1/2 + 1/42)}]/2 = 5 - [3/4 + {5/2 - (21 + 1/42)}]/2 = 5 - [3/4 + {5/2 - 22/42)}]/2 = 5 - [3/4 + {105 - 22/42)}]/2 = 5 - [3/4 + 83/42]/2 = 5 - [63 + 166/84]/2 = 5 - 229/84/2 = 420 - 229/84/2 = 191/84/2 = 191/84 x 2 = 191/168 = 123/168

Question 19

(2.247) 3 + (1.730) 3 + (1.023) 3 -3 x 2.247 x 1.730 x 1.023 / [(2.247) 2 + (1.730) 2 + (1.023) 2 - 2.247 x 1.730 - 1.730 x 1.023 - 2.247 x 1.023] =?

Accepted Answer

We know that, a3 + b3 + c3 - 3abc = (a + b + c) (a2 + b2 + c2 - ab - bc - ca) ⇒ (a + b + c) = (a3 + b3 + c3 - 3abc/a2 + b2 + c2 - ab - bc - ca) ....(i) Given that, (2.247)3 + (1.730)3 + (1.023)3 - 3 x 2.247 x 1.730 x 1.023/ (2.247)2 + (1.730)2 + (1.023)2 - (2.247 x 1.730) - (1.730 x 1.023) - (2.247 x 1.023) = 2.247 + 1.730 x 1.023) - (2.247 x 1.023) = (2.247 + 1.730 + 1.023) [from Eq.(i)] = 5.000 = 5

Question 20

If a 2 + b 2 = a, then the value of a 12 + a 6 + 1 is?

Accepted Answer

a2 + 1 = a ⇒ a + 1/a = 1 On squaring both sides, we get a2 + 1/a2 + 2 = 1 ⇒ a2 + 1/a2 = -1 On cubing both sides, we get (a2 + 1/a2)3 = (-1)3 ⇒ a6 + 1/a6 + 3a2 x 1/a2(a2 + 1/a2) = -1 ⇒ a6 + 1/a6 + 3 x (-1) = -1 Now, a6 + 1/a6 + 1 = 3 As, a12 + a6 + 1 can also be written as a6 + 1/a6 + 1 ∴ a12 + a6 + 1 = 3

Simplification Questions