# Maths questions series #6

## Very important Maths questions for SSC CGL-13 Series#6

Continued from series #5

21. Successive discount of 10% , 20% and 50% will be equivalent to a single discount of-

a. 36%

b. 64%

c. 80%

d. 56%

Ans : Equivalent to a single discount

= [100 – (100 – 10) (100 – 20) (100 – 50)/100*100]%

= [100 – 90*80*50/10000]%

= [100-36]%

= 64%

22. A retailer offers the following discount schemes for buyers on an article-

I. Two successive discounts of 10%

II. A discount of 12% followed by a discount of 8%.

III. Successive discounts of 15% and 5%

IV. A discount of 20%

The selling price will be minimum under the scheme-

a. I

b. II

c. III

d. IV

Ans : From (i) single discount = [10 + 10 – 10*10/100] % = 19%

From (ii) single discount = [12 + 8 -12*8/100] % = 19.04%

From (iii) single discount = [15 + 5 – 15*5/100] % = 19.25%

From (iv) single discount = 20%

We know The S.P. will be minimum under the scheme IV.

23. Of three numbers, the second is thrice the first and the third number is three-fourth of the first. If the average of the three numbers is 114, the largest number is –

a. 72

b. 216

c. 354

d. 726

Ans : Let the first number be x.

We know Second number = 3x and Third number = 3x/4

We know that x + 3x + 3x/4 = 3*114

=> 19x/4 = 342

We know x = 342*4/19 = 72

We know The largest number = 3*72

= 216

24. A car covers 1/5 of the distance from A to B at the speed of 8 km/hour, 1/10 of the distance at 25 km per hour and the remaining at the speed of 20 km per hour. Find the average speed of the whole journey-

a. 12.625 km/hr

b. 13.625 km/hr

c. 14.625 km/hr

d. 15.625 km/hr

Ans : If the whole journey be x km. The total time taken

= (x/5/8 + x/10/25 + 7x/10/20) hrs

= (x/40 + x/250 + 7x/200) hrs

= 25x + 4x + 35x/1000

= 64x/1000 hrs

We know Average speed = x/64x/1000

= 15.625 km/hr

25. The average of 3 numbers is 154. The first number is twice the second and the second number is twice the third. The first number is-

a. 264

b. 132

c. 88

d. 66

Ans : Let the first number be x.

We know Second number = x/2 and Third number = x/4

We know that x + x/2 + x/4 = 3 x 154

=> 7x/4 = 462

We know x = 462*4/7

=264

26. The average salary of all the staff in an office of a corporate house is Rs. 5,000. The average salary of the officers is Rs. 14,000 and that of the rest is Rs. 4,000. If the total number of staff is 500, the number of officers is –

a. 10

b. 15

c. 25

d. 50

Ans : Let the number of officers be x.

We know that 5000*500 = 14000x + 4000(500-x)

We know 2500000 =14000x + 2000000 – 4000x

We know x = 500000/10000

= 50

27. The average marks of 40 students in an English exam are 72. Later it is found that three marks 64, 62 and 84 were wrongly entered as 60, 65 and 73. The average after mistakes were rectified is-

a. 70

b. 72

c. 71.9

d. 72.1

Ans : Correct average

= 40*72 + (64 + 62 +84) – 68 – 65 – 73/40

= 2880 + 210 – 206/40 = 2884/40

= 72.1

28. A and B are two alloys of gold and copper prepared by mixing metals in the ratio 7:2 and 7:11 respectively. If equal quantities of the alloys are melted to form a third alloy C, the ratio of gold and copper in C will be-

a. 5 : 7

b. 5 : 9

c. 7 : 5

d. 9 : 5

Ans : Quantity of gold in A = 7/9* wt. of A

Quantity of gold in B = 7/18* wt. of B

If 1 kg of each A and B are mixed to form third alloys C.

Then quantity of gold in 2 kg C = 7/9 + 7/18

= 7/6 kg

And Quantity of copper in 2 kg C = 2 – 7/6

= 5/6 kg

We know Required ratio = 7/6 : 5/6 = 7 : 5

29. In a laboratory, two bottles contain mixture of acid and water in the ratio 2 : 5 in the first bottle and 7 : 3 in the second. The ratio in which the contents of these two bottles be mixed such that the new mixture has acid and water in the ratio 2 : 3 is-

a. 4 : 15

b. 9 : 8

c. 21 : 8

d. 1 : 2

Ans : Quantity of acid in first bottle = 2/7 x mix.

and Quantity of acid in second bottle = 7/10 x mix.

If x and 1 volumes are taken from I and II bottle respectively to form new mixture.

Then, (2/7 x + 7/10 * 1)/(5x/7 +3/10 * 1) = 2/3

=> 6x/7 21/10 = 10x/7 +6/10

=> 4x/7 = 15/10

We know x = 15/10*7/4 = 21/8

We know Required ratio = x : 1

= 21 : 8

30. A mixture contains 80% acid and rest water. Part of the mixture that should be removed and replaced by same amount of water to make the ratio of acid and water 4 : 3 is-

a. 1/3 rd

b. 3/7 th

c. 2/3 rd

d. 2/7 th

Ans : Let the initial wt. of mixture be 1 kg and x kg of mixture is taken out and replaced by same amount of water.

We know that Amt. of acid/Amt. of water = 0.8 – 0.8x/ (0.2 – 0.2x + x) = 4/3

=> 2.4 – 2.4x = 0.8 + 3.2x

=> 5.6x = 1.6

We know x = 1.6/5.6 = 2/7th part

31. An employer reduces the number of his employees in the ratio 9 : 8 and increases their wages in the ratio 14 : 15. If the original wage bill was Rs. 189,900, find the ratio in which the wage bill is decreased-

a. 20 : 21

b. 21 : 20

c. 20 : 19

d. 19 : 21

Ans : Let the initial number of employees be 9x and the employer gives Rs. 14y as wage to each.

We know that 9x * 14y =18900

We know xy = 150 and The later bill = 8x*15y = 120xy

= 120*150 = 18000

We know Required ratio = 18000 : 18900

= 20 : 21

32. The batting average for 40 innings of a cricketer is 50 runs. His highest score exceeds his lowest score by 172 runs. If these two innings are excluded, the average of the remaining 38 innings is 48 runs. The highest score of the player is-

a. 165

b. 170

c. 172

d. 174

Ans : Let the max. number of runs be x.

We know The lowest score = (x-172)

We know that 40*50 = 38*48 + x + (x-172)

=> 2000 = 1824 + 2x – 172

We know x= 174

33. Walking at 3 km per hour, Pintu reaches his school 5 minutes late. If he walks at 4 km per hour he will be 5 minutes early. The distance of Pintu’s school from his house is-

a. 1 ½ Km

b. 2 Km

c. 2 ½ Km

d. 5 Km

Ans : Distance of the school from the house

= 4*3/(4 – 3) * 5 +5/60 km

= 12*1/6

=2 km

34. Nitin bought some oranges at Rs. 40 a dozen and an equal number at Rs.30 a dozen. He sold them at Rs. 45 a dozen and made a profit of Rs.480. The number of oranges, he bought, was-

a. 48 dozen

b. 60 dozen

c. 72 dozen

d. 84 dozen

Ans : Let the number of oranges bought be x.

We know 45x/12 – 70x/24 = 480

=> (45-35)/12 = 480

=> x = 480*12/10 = 576

= 48*12 = 48 dozen

35. A man buys two chairs for a total cost of Rs.900. By selling one for 4/5 of its cost and the other for 5/4 of its cost, he makes a profit of Rs.90 on the whole tr

36. By selling 100 oranges, a vendor gains the selling price of 20 oranges. His gain per cent is-

a. 20

b. 25

c. 30

d. 32

Ans : Let the S.P. of 100 Oranges be Rs. x.

We know S.P. of 20 oranges = x/100 *20 = Rs. x/5

We know C.P. of 100 oranges = x – x/5

= Rs. 4x/5

We know Reqd. Profit % = x/5 * 100*5/4x %

= 25%

37. 60% of the cost price of an article is equal to 50% of its selling price. Then the percentage of profit or loss on the cost price is-

a. 20% loss

b. 16 2/3% profit

c. 20% profit

d. 10% loss

Ans : Let the cost price be Rs. 100.

We know that S.P * 50/100 = 100*16/100

We know S.P = 60*100/50

= Rs. 120

We know Reqd. % profit = (120 – 100)% = 20%

38. Maninder bought two horses at Rs.40,000 each. He sold one horse at 15% gain, but had to sell the second horse at a loss. If he hand suffered a loss of Rs.3,600 on the whole tr

39. A fruit-seller buys x guavas for Rs.y and sells y guavas for Rs. x. If x>y, then he made-

a. x2 – y2 / xy % loss

b. x2 – y2 / xy % gain

c. x2 – y2 / y2 % loss

d. x2 – y2 / y2 * 100% gain

Ans : C.P. of 1 guava = Rs. y/x [x>y]

and S.P. of 1 guava = Rs. x/y

We know Reqd. Gain% = x/y – y/x/ y/x * 100%

= x2 – y2 /y2 *100%

40. A jar contain 10 red marbles and 30 green ones. How many red marbles must be added to the jar so that 60% of the marbles will be red?

a. 25

b. 30

c. 35

d. 40

Ans : Let after adding x red marbles, the red marbles with be 60% of the total.

We know that (10+x)/ (10 + x) + 30 = 60/100

=> 10 + x/40 + x = 3/5

=> 50 + 5x = 120 + 3x

x = 70/2 = 35

41. If a number multiplied by 25% of itself gives a number which is 200% more than the number, then the number is –

a. 12

b. 16

c. 35

d. 24

Ans : Let the number be x.

We know that x*25x/100 = x + 200x/100

=> x2/4 = 3x

=> x2 – 12x = 0

=> x – 12 = 0

We know x = 12

42. The value of an article depreciates every year at the rate of 10% of its value. If the present value of the article is Rs.729, then its worth 3 years ago was-

a. Rs.1250

b. Rs.1000

c. Rs.1125

d. Rs.1200

Ans : Let the worth 3 years ago be Rs. x.

We know that 729 = x (1 – 10/100)3

=> 729 = x*9*9*9/10*10*10

We know x = Rs. 1000

43. The price of onions has been increased by 50%. In order to keep the expenditure on onions the same the percentage of reduction in consumption has to be-

a. 50%

b. 33 1/3%

c. 33%

d. 30%

Ans : Reqd. Percentage of reduction

= 50*100/(100 + 50) %

= 5000/150 %

= 33 1/3%

44. A took two lo

Ansaction. The cost of the lower priced chair is-

a. Rs.360

b. Rs.400

c. Rs.420

d. Rs.300

[showhide type=“q35” ] Ans : Let the cost price of 1 chair be Rs. x.

We know C.P. of other chair = Rs. (900-x)

We know that 4/5 x + 5/4 (900-x) = 900 + 90

=> 4/5 x +1125 -5x/4 = 990

We know 9x/20 = 135

We know x = 135*20/9

= Rs. 300

We know C.P. of the lower priced chair is Rs. 300.

45. If a regular polygon has each of its angles equal to 3/5 times of two right angles, then the number of side is-

a. 3

b. 5

c. 6

d. 8

Ans : If the number of sides a regular polygon be n.

Then (2n-4)/n = 2*3/5

=> (2n – 4)*5 = 6n

We know n = 5

46. A square is of area 200 sq. m. A new square is formed in such a way that the length of its diagonal is ?2 times of the diagonal of the given square. The the area of the new square formed is-

a. 200?2 sq.m

b. 400?2 sq.m

c. 400 sq.m

d. 800 sq.m

Ans : Length of the diagonal of Ist square

= ?2*200

= 20 m

We know Length of the diagonal of new square = 20?2m

We know Area of the new square = ½*(20.?2)2 = 400 sq. m

47. A motor-boat can travel at 10 km/hr in still water. It travelled 91 km downstream in a river and then returned to the same place, taking altogether 2 hours. Find the rate of flow of river-

a. 3 km/hr

b. 4 km/hr

c. 2 km/hr

d. 5 km/hr

Ans : Let the rate of flow of river be x km/hr.

We know that 91/(10 + x) + 91/(10 – x) = 20

=> 91(10 – x + 10 + x)(10 + x) (10 – x) = 20

=> 91*20 = 20(100 – x2)

=> x2 = 9 = (3)2

We know x = 3 km/hr

48. A man driving at 3/4th of his original speed reaches his destination 20 minutes later than the usual time. Then the usual time is-

a. 45 minutes

b. 60 minutes

c. 75 minutes

d. 120 minutes

Ans : Let the original speed be x km/hr and the usual time be y hours.

We know that x * y = ¾ x(y+1/3)

We know 4y = 3y + 1

We know y = 1 hr = 60 minutes

50. A motor-boat, travelling at the same speed, can cover 25 km upstream and 39 km downstream in 8 hours. At the same speed, it can travel 35 km upstream and 52 km downstream in 11 hours. The speed of the stream is –

a. 2 km/hr

b. 3 km/hr

c. 4 km/hr

d. 5 km/hr

[showhide type=“q50” ] Ans : Let the speeds of motor boat and the stream be x and y km/hr respectively.

We know that 39/x + y + 25/x – y = 8 …(1)

and 52/x + y + 35/x – y = 11 …(2)

Solving equations (1) and (2), we get-

We know 100 – 105/x – y = 32 – 33

We know x – y = 5

and x + y = 13

We know y = 4 km/hr