**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.

**1) The side of square and its area is in the ratio of 1 : 11. Find the perimeter of the square?**

a) 48 cm

b) 40 cm

c) 44 cm

d) None of these

**2) A boat can cover a distance of 240 kms upstream in 16 hours while the same distance downstream in 10 hours. Find the speed of stream?**

a) 4.5 kmph

b) 3.5 kmph

c) 5 kmph

d) 9 kmph

**3)** **LCM of two numbers 525 and their HCF is 20. If one number is 50, then the other number is**

a) 125

b) 175

c) 280

d) 210

**4)** **Which of the following relation is true?**

a) Mode = Median – Mean

b) Mode = 3Median + 2Mean

c) Mode = 3Median – Mean

d) Mode = 3Median – 2Mean

**5) Find the average of first 20 multiples of 13**

a) 136.5

b) 137.25

c) 136.25

d) 137.5

**6)** **A Shopkeeper sells some bananas. He sells 37.5% of bananas and still, he has 500 bananas left. Originally, he had:**

a) 750 bananas

b) 850 bananas

c) 872 bananas

d) 800 bananas

**7) Find the value of (sec α – tan α) ^{2?}**

a) cotα

b) (1 – sin α)/(1 + sin α)

c) sin α × cosec α

d) 0

**8) If a + b = 18 and ab = 54, then find the value of a – b.**

a) 6

b) 5√2

c) 7

d) 6√3

**9) In the circle given below, chord AB** **is extended to point D to match the tangent DE. If AB = 18 cm and BD = 6 cm, then find the length of DE?**

a) 8 cm

b) √27cm

c) 12 cm

d) 10 cm

**10) Breadth of the cuboidal box is half its length and one fifth its height. Find the lateral surface area of the cuboidal box if its volume is 58,320 cm ^{2}.**

a) 11,016 cm^{2}

b) 10,208 cm^{2}

c) 9,720 cm^{2}

d) None of these

**Answers :**

**1) Answer: C**

Let the side of square = a cm

According to the question

a : a^{2} = 1 : 11

a = 11 cm

Side = 11 cm

Perimeter of square = 4 x side of square

Perimeter of square = 44 cm.

**2) Answer: A**

Let the upstream and downstream speed be ‘Us’ and ‘Ds’ respectively.

Speed = Distance/time

Us = 240/16 = 15 kmph

Ds = 240/10 = 24 kmph

Let ‘U’ and ‘V’ be the speed of boat in still water and speed of current respectively.

U + V = 24 ……………..(1)

U – V = 15 …………..(2)

Solving eq (1) and eq (2)

Speed of stream (V) = 4.5 kmph

**3) Answer: D**

LCM × HCF = 1^{st} Number × 2^{nd} Number

525 × 20 = 50 × x

x = (525 × 20)/50 = 210

**4) Answer: D**

We know that:

Mode = 3Median – 2Mean

**5) Answer: A**

First 20 numbers which are multiples of 13= 13, 26, 39,…..260

Average of those 20 numbers= (13+26+39+….+260)/20

= 13(1+2+3…..+20)/20 -(1)

Sum of first n natural numbers=n(n+1)/2

Apply the above formula then

(1)=> 13(20*21)/40= 136.5

**6) Answer: D**

Percentage of Left out bananas = 100% – 37.5% = 62.5%

As per the question,

100/62.5×500 = 800 bananas

**7) Answer: B **

(sec α – tan α)^{2} = ((1/cos α) – (sin α/cos α))^{2}

= ((1 – sin α/cos α))^{2}

= (1 – sin α)^{2}/cos^{2} α

= (1 – sin α)^{2}/(1 – sin^{2} α) [∴sin^{2} α + cos^{2} α = 1]

= [(1 – sin α)(1 – sin α)/(1 – sin α)(1 + sin α)]

= [(1 – sin α)/(1+sin α)]

**8) Answer: D**

(a + b)^{2} = a^{2} + b^{2} + 2ab

18^{2} = a^{2} + b^{2} + 2 × 54

a^{2} + b^{2} = 324 – 108

a^{2} + b^{2} = 216

Now (a – b)^{2} = a^{2} + b^{2} – 2ab

(a – b)^{2} = 216 – 2 × 54

(a – b)^{2} = 216 – 108

(a – b)^{2} = 108

(a – b) = √108

(a – b) = 6√3

**9) Answer: C**

From the theorem,

DE^{2 }= DB × DA = 6 × 24

DE^{2 }= 144

DE = 12 cm

**10) Answer: C**

Let the breadth of the cuboidal box = x cm

Length of the cuboidal box = 2x cm

Height of the cuboidal box = 5x cm

So the volume of the box = x × 2x × 5x = 58320

10x^{3} = 58320

x^{3} = 5832, x = 18

So the length, breadth and the height of the box are 36 cm, 18 cm and 90 cm respectively.

So the lateral surface area of the box = 2 × 90 × (18 + 36) = 9,720 cm^{2}