SSC CGL Mains Quantitative Aptitude Questions : 27th July_00.1
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SSC CGL Mains Quantitative Aptitude Questions : 27th July

Dear aspirants,

As you all know, the upcoming months are lined up with various important exams like SSC CGL Mains 2018, so we are here to help you with the subject that is common to all of the given exams. We are providing daily quantitative aptitude quizzes, practice which will help you to score good marks in this section. We aim to provide the best study material to our readers with exam level questions to help them get used to the recent pattern. Attempt this quiz and check your preparation.

Q1. The sum of the ages of a mother and son is 45 yrs, Five years ago the product of their ages was 4 times the mother’s age at that time. The present age of the mother and son, respectively are
एक माँ और बेटे की उम्र का योग 45 वर्ष है, पाँच वर्ष पहले उनकी उम्र का गुणनफल उस समय माँ की उम्र का 4 गुना था। माँ और बेटे की वर्तमान आयु क्रमशः क्या है ?
(a) 36 yrs, 9 yrs /36 वर्ष, 9 वर्ष
(b) 39 yrs, 6 yrs /39 वर्ष, 6 वर्ष
(c) 42 yrs, 12 yrs/ 42 वर्ष, 12 वर्ष
(d) 25 yrs, 10 yrs /25 वर्ष, 10 वर्ष
Ans.(a)
Sol. Let the present ages of mother and son be x years and (45 – x) years respectively.
Then, (x-5)(45-x-5)=4(x-5)
⇒x²-41x+180=0
⇒x=36
∴ The present ages of mother and son are 36 yrs & 9 yrs. respectively.

Q2. If we divide 3150 into four parts such that half of the first part, a third of the second part and a fourth of the third part are equal to one-twelfth of the fourth part. Then find the largest part?
यदि हम 3150 को चार भागों में विभाजित करते हैं जैसे कि पहले भाग का आधा भाग, दूसरे भाग का एक तिहाई और तीसरे भाग का एक चौथाई भाग, चौथे भाग के एक-बारहवें हिस्से के बराबर होता है। फिर सबसे बड़ा हिस्से बताएं?
(a) 1800
(b) 1200
(c) 600
(d) None of these /इनमें से कोई नहीं
Ans.(a)
Sol. Let the four parts into which 3150 is divided are a, b, c and d.
⇒a/2=b/3=c/4=d/12=k
Then a = 2k, b = 3k, c = 4k and d = 12k
As a + b + c + d = 3150
⇒(2k+3k+4k+12k)=3150
⇒21k=3150
⇒k=150
Hence the four parts are 300, 450, 600, 1800
So, the largest part is 1800

Q3. A hemispherical bowl of internal radius 36 cm, contains a liquid. This liquid is to be filled into small cylindrical bottles of diameter 12 cm and height 3 cm, then the number of bottles necessary to empty the bowl is
36 सेमी आंतरिक त्रिज्या के एक गोलार्द्ध कटोरे में एक तरल होता है. इस तरल को 12 सेमी व्यास और 3 सेमी ऊंचाई की छोटी बेलनाकार बोतलों में भरा जाना है, फिर कटोरे को खाली करने के लिए आवश्यक बोतलों की संख्या कितनी है?
(a) 288
(b) 252
(c) 144
(d) 154
Ans.(a)
Sol.
No.of bottles=(Volume of hemisphere)/(Volume of cylinder)
=(2/3 πR³)/(πr² h)=2/3×(36×36×36)/(6×6×3)
= 288

Q4. The present population of a kheda village is 5500. If the number of males increases by 11% and the number of females increases by 20%, then the population will becomes 6330. What is the present population of females in the kheda village?
खेडा गाँव की वर्तमान आबादी 5500 है. यदि पुरुषों की संख्या 11% बढ़ा दी जाती है और महिलाओं की संख्या 20% तक बढ़ा दी जाती है, तो आबादी 6330 हो जाती है. तो खेडा गाँव में महिलाओं की वर्तमान आबादी कितनी है?
(a) 3000
(b) 3500
(c) 2500
(d) 2000
Ans.(c)
Sol. Let the present population of male and female be x and y respectively.
ATQ, x+y=5500
And 111x/100+120y/100=6330
⇒111x+120y=633000
From equations (i) and (ii), we get
Y = 2500

Q5. In the given figure, ABCD is a square. EFGH is a square formed by joining mid points of sides of ABCD. LMNO is a square formed by joining mid points of sides of EFGH. A circle is inscribed inside EFGH. If area of circle is 38.5 cm², then what is the area (in cm²) of square ABCD?
दी गयी आकृति में, ABCD एक वर्ग है. EFGH, ABCD की भुजाओं के मध्य बिंदुओं को मिलाकर बना एक वर्ग है. LMNO, EFGH की भुजाओं के मध्य बिंदुओं को मिलाकर बना एक वर्ग है. EFGH के अंदर एक वृत्त अंकित किया जाता है। यदि वृत्त का शेत्रफल 38.5 सेमी² है, तो वर्ग ABCD का शेत्रफल (सेमी²) में क्या है?
SSC CGL Mains Quantitative Aptitude Questions : 27th July_50.1
(a) 98
(b) 196
(c) 122.5
(d) 171.5

Ans.(B)
Sol.

From the following diagram , area of circle given is 38.5, hence radius is
22/7×r^2=38.5
(r=3.5)
So,
(LM=2*r=7=MN=NO)
MO is diagonal of square MNOL
(MO=7√2=EF)
EF is side of second square EFHG
EH is diagonal of EFHG
EH = √2×7√2=14
SIDE of square ABCD=14
therefore
(Area ABCD=(14)²=196)

Q6. ABCDEF is a regular hexagon of side 12 cm. What is the area (in cm²) of the triangle ECD?
ABCDEF, भुजा 12 सेमी का एक नियमित षट्भुज है। त्रिभुज ECD का क्षेत्रफल (cm² में) क्या है?
(a) 18√3
(b) 24√3
(c) 36√3
(d) 42√3
Ans.(C)
Sol.GIVEN below is a regular hexagon and
SSC CGL Mains Quantitative Aptitude Questions : 27th July_60.1
A regular hexagon has 6 equilateral triangle inside it as shown in diagram
and area of ∆ECD =area of (1/2)×2 equilateral triangle
∆ECD =area of 1equilateral triangle
Area of ∆ECD = √3/4×12×12
(∆ECD=36√3)

Q7. PQRS is a square whose side is 16 cm. What is the value of the side (in cm) of the largest regular octagon that can be cut from the given square?
PQRS एक वर्ग है जिसकी भुजा 16 सेमी है। दिए गए वर्ग से काटे जाने वाले सबसे बड़े नियमित अष्टकोण की भुजाओं (सेमी में) का मान क्या है?
(a) 8 – 4√2
(b) 16 + 8√2
(c) 16√2 – √16
(d) 16 – 8√2
Ans.(C)
Sol. Inside a square of side 16, aoctagone of side y is constructed as shown below
SSC CGL Mains Quantitative Aptitude Questions : 27th July_70.1
With respect to figure
PS=2x + y = 16
x = ((16–y)/2)
IN triangle PTU
x²+x²=y²
PUT VALUE OF x
2(((16–y)/2))²=y²
256 + y² –32y – 2y²
y² + 32y – 256 = 0
y = (–32 ± √((32)²+ 4 × 256))/2
y = –16 ±√2×16
(y=16√2–16) (negative values neglected as side cannot be negative )

Q8. Rajkamal’s factory kept increasing its output by the same percentage every year. Find the percentage increase every year if it is known that his output is doubled after two years.
राजकमल के कारखाने का उत्पादन हर साल समान प्रतिशत से बढ़ता है। अगर यह ज्ञात हो कि उसका उत्पादन दो साल बाद दोगुना हो जाता है तो हर साल प्रतिशत वृद्धि ज्ञात करें।
(a) 100√2%
(b) 100(√2+1)%
(c) 100(√2-1)%
(d) 50(√3-1)%
Ans.(c)
Sol. Let the output be x and percentage be a.
Then,
∴x×(1+a/100)²=2x
⇒1+a/100=√2
⇒a/100=√2-1
⇒a=(√2-1)×100%
=100(√2-1)%

Q9. Two pipes can fill a tank in 8 h and 12 h respectively whereas an escape pipe can empty it in 6 h. If the three pipes are opened at 1 pm, 2 pm and 3 pm respectively, at what time will the tank be filled?
दो पाइप एक टंकी को क्रमशः 8 घंटे और 12 घंटे में भर सकते हैं जबकि एक रिसाव पाइप इसे 6 घंटे में खाली कर सकता है. यदि तीन पाइप क्रमश: दोपहर 1 बजे, 2 बजे और 3 बजे खोले जाते हैं, तो टंकी कितने बजे तक भरेगी?
(a) 8 am next day/अगले दिन सुबह 8 बजे
(b) 7 am next day /अगले दिन सुबह 7 बजे
(c) 5 am next day/अगले दिन सुबह 5 बजे
(d) 7.30 am next day/अगले दिन सुबह 7:30 बजे
Ans.(b)
Sol. Till 3 pm, part of tank filled
=2/8+1/12=1/4+1/12=4/12=1/3 part
∴ Remaining part = 1-1/3=2/3
Now, let x be the time taken by all three pipes work to fill the remaining part of tank
(1/8+1/12-1/6)x=2/3
⇒(6+4-8)x/48=2/3
⇒2x/48=2/3⇒x=16 hour
Duration of time = 16 hours
And Tank will be filled at = 3pm + 16 hours
= 7 am next day

Q10. The total monthly sales of two companies A and B are in the ratio 2 : 3 and their total monthly expenditures are in the ratio 3 : 4. Find the ratio of the profits of two companies given that company A’s profit is equal to one fifth of its sales.
दो कंपनी A और B की कुल मासिक बिक्री 2: 3 के अनुपात में है और उनका कुल मासिक व्यय 3: 4 के अनुपात में है. यदि कंपनी A का लाभ उसकी बिक्री के पांचवे हिस्से के बराबर है तो दोनों कंपनियों के लाभ के अनुपात का पता लगाएं।
(a) 6 : 13
(b) 8 : 15
(c) 13 : 6
(d) 15 : 8
Ans.(a)
Sol. Let the total monthly sales of companies A and B be Rs. 2x and Rs. 3 x and their total monthly expenditure be Rs. 3y & Rs. 4y.
Given that A’s profit = 1/5 of sales = 2x/5
∴2x-3y=1/5(2x)
⇒4/5 (2x)=3y⇒y=8/15 x
Profit of company
B = 3x-4y=3x-4×8/15 x=13x/15
Hence the ratio of the profits of the two companies are
2/5 x:13x/15=6∶13

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