# Quantitative Aptitude Questions Daily Quiz Day – 25

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153 Dear Aspirants, Our SSC Crackers team is providing a new series of Quantitative Aptitude Questions for Upcoming Exam so the aspirants can practice it on a daily basis. These questions are framed by our skilled experts after understanding your needs thoroughly. Aspirants can practice these new series questions daily to familiarize with the exact exam pattern and make your preparation effective.

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1) The full mark for a paper is 300. The break-up of the marks into theory (X), practical (Y) and (Z) project, which are the three components of evaluation is 6 : 5 : 4. In order to pass one has to score at least 40%, 50% and 50% respectively in XYZ and 60% in aggregate. The marks scored by four students A, B, C and D are shown in the given bar graph. What is the average mark of B, C and D in project?

a) 30

b) 40

c) 50

d) 60

2) Find the value of:

(68.61 × 68.61 – 12. 39 × 12.39) ÷ 56.22

a) 98

b) 56.22

c) 77

d) 81

3) A person bought a watch and a tube light for Rs. 700 and Rs. 170 respectively and sold them to another person at gain of 12% and 10% respectively. Find his overall gain percentage.

a) 12%

b) 13%

c) 11.60%

d) 12.5%

4) If ‘s’ is a natural number, then the largest number dividing (s3 – s) is:

a) 6

b) 7

c) 4

d) 5

5) The perimeter of two similar triangles XYZ and LMN are 140 cm and 119 cm respectively. What is the length of XY, if LM = 17 cm?

a) 18 cm

b) 20 cm

c) 17 cm

d) 21 cm

6) The average speed of a bus is 3/4th of the average speed of a train. The train covers 240 km in 12 h. How much distance will the bus cover in 7 h?

a) 210 km

b) 15 km

c) 20 km

d) 105 km

7)  The full mark for a paper is 300. The break-up of the marks into theory (X), practical (Y) and (Z) project, which are the three components of evaluation is 6 : 5 : 4. In order to pass one has to score at least 40%, 50% and 50% respectively in XYZ and 60% in aggregate. The marks scored by four students A, B, C and D is shown in the given bar graph. Who among the person given above has failed in only one of the subjects but has qualified in aggregate?

a) B

b) D

c) C

d) A

8) If x is an integer, then which of the following is a solution of 2/3 + 1/3 (x – 1) > 0

a) -3

b) -4

c) -2

d) 2

9) The simplified value of

(0.5 x 0.5 x 0.5 + 0.2 x 0.2 x 0.2 + 0.3 x 0.3 x 0.3 – 3x 0.5 x 0.2 x 0.3) ÷ (0.5 x 0.5 + 0.2 x 0.2 + 0.3 x 0.3 – 0.5 x 0.2 – 0.2 x 0.3 – 0.5 x 0.3) = ?

a) 1

b) 0.6

c) 0.03

d) 0.4

10) 812.5 x 94.5 ÷ 34.8 = 9?

a) 7.1

b) 9.4

c) 4.7

d) 4.5

Full marks = 300

The break-up of the marks into theory (X), practical (Y) and (Z) project, which are the three components of evaluation is 6 : 5 : 4.

So, 300 ÷ (6 + 5 + 4) = 20

Break-up of marks in:

Theory (X) = 6 x 20 = 120

Practical (Y) = 5 x 2 = 100

Project (Z) = 4 x 20 = 80

Marks scored by the persons in project:

B = 30, C = 60, D = 30

Average = (30 + 60 + 30) ÷ 3 = 40

(68.61 × 68.61 – 12. 39 × 12.39) can be written in the form of (a + b) (a – b)

= (68.61 + 12.39) (68.61 – 12.39)

= 81 x 56.22

(68.61 × 68.61 – 12. 39 × 12.39) ÷ 56.22 = (81 x 56.22) x 56.22 = 81

Profit on watch = 12% of 700 = Rs. 84

Profit on tube light = 10% of 170 = Rs. 17

Total CP = (700 + 170) = 870

Total profit = (84 + 17) = 101

101 is 11.60% of 870

(s3 – s) → (23 – 2) = 6 is the largest natural that divides (s3 – s) for every number ‘s’.

Perimeter of triangle XYZ = 140 cm

Perimeter of triangle LMN = 119cm

XY = ?

LM = 17 cm

In similar triangles, ratios of the perimeter of the triangles are equal to the ratio of their respective sides.

So,

140/119 = XY/17

Or, XY = 20 cm

Average speed of the train = 240/12 = 20 km/h

Average speed of bus = 3/4 × 20 = 15 km/h

Distance = 7 × 15 = 105 km

Full marks = 300

The break-up of the marks into theory (X), practical (Y) and (Z) project, which are the three components of evaluation is 6 : 5 : 4.

So, 300 ÷ (6 + 5 + 4) = 20

Break-up of marks in:

Theory (X) = 6 x 20 = 120   Pass Marks = 40% of 120 = 48

Practical (Y) = 5 x 2 = 100    Pass Marks = 50% of 100 = 50

Project (Z) = 4 x 20 = 80      Pass Marks = 50% of 80 = 40

Aggregate % in overall = 60% of 300 = 180

Score of:

A = 180

B = 130

C = 210

D = 180

D has scored 30 in practical (Z), whereas the passing marks is 40.

Hence, ‘D’ is the person who has failed in only one of the subjects but has qualified in aggregate.

a) -3 → 2/3 + 1/3 (x – 1) > 0

L.H.S = 2/3 + 1/3 (-3 – 1) is not greater than 0

b) -4 → 2/3 + 1/3 (x – 1) > 0

L.H.S = 2/3 + 1/3 (-4 – 1) is not greater than 0

c) -2 → 2/3 + 1/3 (x – 1) > 0

L.H.S = 2/3 + 1/3 (x – 1) is not greater than 0

d) 2 → 2/3 + 1/3 (x – 1) > 0

L.H.S =  2/3 + 1/3 (x – 1) is greater than 0

Given expression: (0.5 x 0.5 x 0.5 + 0.2 x 0.2 x 0.2 + 0.3 x 0.3 x 0.3 – 3x 0.5 x 0.2 x 0.3) ÷ (0.5 x 0.5 + 0.2 x 0.2 + 0.3 x 0.3 – 0.5 x 0.2 – 0.2 x 0.3 – 0.5 x 0.3)

It is the simplified form of:

(a3 + b3 + c3 – 3abc) ÷ (a2 + 2 + c2 – ab – bc – ca) = a + b + c

So, a + b + c = 0.5 + 0.2 + 0.3 = 1