# Quantitative Aptitude Questions Daily Quiz Day – 24

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137 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 sum of five consecutive even numbers of set-A is 280. What is the sum of five consecutive numbers of a different set, whose lowest number is 71 less than the double of the lowest number of set-A?

a) 182

b) 165

c) 172

d) 175

2)  A man bought some apples for Rs. 224 at the rate of Rs. 8 per apple. If 3 apples get rotten and he sold the remaining apples at the rate of Rs. 11.2 per apple, find his profit/loss percentage.

a) 16.66%

b) 20%

c) 25%

d) 33.33%

3) A ladder rests against a wall with its lower end at a distance x from the wall and its upper end at a height of 2x above the floor. If the lower end slides through a distance y away from the wall, from its earlier position then by how much distance does the upper end slide? a) b) c) d) None of these

4) What is the value of sin2A + sin2(120 + A) + sin2(120 – A)?

a) 3/2

b) -1/2

c) 1/2

d) None of these

5) ABCD is a rhombus, in which diagonal AC and BD is 8 cm and 15 cm respectively. Find the area of the rhombus.

a) 60 cm2

b) 72 cm2

c) 90 cm2

d) 78 cm2

6) In the given circle with center O, angle QPR measures 48o. Find the measure of angle ORT. a) 138o

b) 156o

c) 128o

d) 146o

7) If sec β = 3/2, find the value of cot2β – sin2β. (Given: 0o < β < 90o)

a) 11/23

b) 23/45

c) 11/45

d) 1/4

8) Minimum and maximum value of sin(sin x) are

a) Do not exist

b) -1, 1

c) sin (x) -1 , sin(x) + 1

d) –sin (1), sin(1)

9) The perimeter of five squares are 24 cm, 32 cm, 40 cm, 80 cm and 100 cm respectively. The perimeter of another square equal in area to sum of the areas of these squares is:

a) 152 cm

b) 124 cm

c) 140 cm

d) 70 cm

10) ABCD is a trapezium, such that AB = CD and AD || BC, AD = 6 cm, BC = 8 cm. If area of ABCD is 70 cm2 then CD is –

a) √101 cm

b) √112 cm

c) √104 cm

d) √113 cm

The average of these five consecutive even numbers = 280 / 5 = 56

So 56 is the middle number of set A

Lowest number of set A = 56 -4 = 52

Let the other set be set B.

Lowest number of set B = (2 x 52) – 71 = 33

So set B = { 33, 34, 35, 36, 37}

Sum of set B = 175.

Number of apples bought = 224 ÷ 8 = 28

Remaining number of apples sold = 28 – 3 = 25

Selling price of 25 apples = 25 × 11.2 = Rs. 280

Cost price of 28 apples = Rs. 224

Required profit percentage = {(280 – 224)/224} × 100 = 25%  Area of rhombus = 1/2 × product of diagonals

= 0.5 × 8 × 15

= 60 cm2

Measure of angle QSR = Measure of angle QPR = 48 (Angles in the same segment)

Measure of angle QOR = 2 × Measure of angle QSR = 2 × 48o = 96o (Angle formed at the center is twice of the angle formed at the circumference subtended by the same arc)

Since OQ and OR are radii of the same circle, measures of angle OQR and ORQ in triangle OQR are equal (angles opposite to equal sides of a triangle are equal).

As per angle sum property of triangles, measure of angle ORQ = 42o

Consequently, measure of angle ORT = 180o – 42o = 138o (Linear pair)

If sec β = 3/2, cos β = 2/3. Then, cos2β = 4/9

Thus, sin2β = 1 – cos2β = 1 – 4/9 = 5/9

Then, cot2β = cos2β/sin2β = (4/9)/ (5/9) = 4/5

Thus, cot2β – sin2β = 4/5 – 5/9 = 11/45

We know that, -1 ≤ sin(nx) ≤ 1

= sin (-1) ≤ sin (sin x) ≤ sin (1)

= –sin 1 ≤ sin (sin x) ≤ sin 1; [sin(-θ) is same as –sin θ]

Therefore, Minimum value is –sin 1 and maximum is sin 1

Perimeter of the first square = 24 cm.

∴ side of the first square = 24/4 = 6 cm.

∴ Area of the first square = 62 = 36 cm.

Perimeter of the second square = 32 cm.

∴ side of the second square = 32/4 = 8 cm.

∴ Area of the second square = 82 = 64 cm.

Perimeter of the third square = 40 cm.

∴ side of the third square = 40/4 = 10 cm.

∴ Area of the third square = 102 = 100 cm.

Perimeter of the fourth square = 80 cm.

∴ side of the fourth square = 80/4 = 20 cm.

∴ Area of the fourth square = 202 = 400 cm.

Perimeter of the fifth square = 100 cm.

∴ side of the fifth square = 100/4 = 25 cm.

∴ Area of the fifth square = 252 = 625 cm.

∴ Area of the bigger square,

⇒ (36 + 64 + 100 + 400 + 625)

⇒ 1225 sq. cm.

∴ side of the bigger square(a) = √1225 = 35 cm.

∴ Perimeter of the bigger square = 4 × 35 = 140 cm.

We know that,

Area of trapezium = (1/2) × Height × Sum of the parallel sides

⇒ 70 = (1/2) × Height × (6 + 8)

⇒ Height of the trapezium = 10 cm

By Pythagoras theorem,

CD2 = DE2 + EC2

AB2 = AF2 + BF2

(∵ AB = CD and AF = DE = h = 10 cm)

⇒ EC = BF and EC = BF = (BC – AD)/2 = (8 – 6)/2 = 2/2 = 1 cm

CD2 = DE2 + EC2

⇒ CD2 = 102 + 12

⇒ CD = √101 cm