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SSC CGL Maths Mains Questions : 29th June

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.

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Q1. (180 × 15 – 12 × 20)/(140 × 8 + 2 × 55) = ?
(a) 1/7
(b) 4/5
(c) 2
(d) 4
Ans.(c)
Sol. Given expression = (2700 – 240)/(1120 + 110)=2460/1230 = 2.

Q2. Given that (1² + 2² + 3² + ………… + 10²) = 385, the value of (2² + 4² + 6² + ………… + 20²) is equal to
दिया गया है कि (1² + 2² + 3² + ………… + 10²) = 385, तो (2² + 4² + 6² + ………… + 20²) का मान किसके बराबर है-
(a) 770
(b) 1155
(c) 1540
(d) (385)²
Ans.(c)
Sol. (2² + 4² + 6² + . . . . . . . . . + 20²) = 2² (1² + 2² + 3² + . . . . . . . . . + 10²)
= 4 × 385 = 1540.

Q3. A milkman has 3 jars containing 57 litres, 129 litres and 177 litres of pure milk respectively. A measuring can, after a different number of exact measurements of milk in each jar, leaves the same amount of milk unmeasured in each jar. What is the volume of the largest such can?
एक दूधवाले के पास क्रमशः 57 लीटर, 129 लीटर और 177 लीटर शुद्ध दूध के 3 जार है। प्रत्येक जार में दूध के अलग अलग एक सटीक माप के बाद, प्रत्येक जार में दूध की समान मात्रा शेष बचती है। सबसे बड़े जार की मात्रा कितनी है?
(a) 12 litres/लीटर
(b) 16 litres/लीटर
(c) 24 litres/लीटर
(d) None of these/इनमें से कोई नहीं
Ans.(c)
Sol. Required volume = [H.C.F. of (129 – 57), (177 – 129) and (177 – 57)] litres
= (H.C.F. of 72, 48 and 120) litres = 24 litres

Q4. If x = (a/b) + (b/a), y = (b/c) + (c/b) and z = (c/a) + (a/c), then what is the value of xyz – x² – y² – z²?
यदि x = (a/b) + (b/a), y = (b/c) + (c/b) and z = (c/a) + (a/c) है,तो xyz – x² – y² – z² का मान कितना है?
(a) – 4
(b) 2
(c) – 1
(d) – 6
Ans.(A)
Sol. GIVEN
x=a/b+b/a, y=b/c+c/b, &z=c/a+a/c
put a = b = c = 1
x = y = z will come out as 2
put in eqn. to find xyz – (x² + y² + z²)
⇒ 2 × 2 × 2 – (4 + 4 + 4)
⇒ – 4

Q5. A confectionery shopkeeper sold a toothpaste at a profit of 17.5%. If he had bought it at 8% less and sold it at 30% profit, he would have earned Rs. 11.55 more as profit. Cost price of the toothpaste is
एक दुकानदार ने 17.5% के लाभ पर एक टूथपेस्ट बेचा। यदि वह इसे 8% कम पर खरीदता और 30% लाभ पर बेचता, तो वह लाभ के रूप में 11.55 रुपये अधिक अर्जित करता। टूथपेस्ट का लागत मूल्य कितना है
(a) Rs. 550 /रूपये
(b) Rs. 675/रूपये
(c) Rs. 750/रूपये
(d) Rs. 1475/रूपये
Ans.(a)
Sol. Let cost price of the toothpaste = Rs. x.
So, x×(117.5/100)+11.55=x×(92/100)×(130/100)
⇒x=11.55/((0.92×130-117.5))×100
=Rs.550

Q6. Anu is a working partner and Bimla is a sleeping partner in a business. Anu puts in Rs. 5000 and Bimla puts in Rs. 6000. Anu receives 12.5% of the profit for managing the business and the rest is divided in proportion to their capitals. What does each get out of a profit of Rs. 880?
एक बिजनेस में अनु एक कार्यकारी साझेदार है और बिमला एक कार्यकारी साझेदार नहीं है। अनु 5000 रुपये का निवेश करती है और बिमला 6000 रुपये का निवेश करती है। अनु को व्यवसाय के प्रबंधन के लिए लाभ का 12.5% प्राप्त होता है और शेष उनकी निवेश की राशी के अनुपात में विभाजित होता है। 880 रुपये के लाभ में प्रत्येक को कितना लाभ प्राप्त होता है?
(a) Rs. 400 and Rs. 480 / 400 रूपये और 480 रूपये
(b) Rs. 460 and Rs. 420 /460 रूपये और 420 रूपये
(c) Rs. 450 and Rs. 430 /450 रूपये और 430 रूपये
(d) Rs. 470 and Rs. 410 /470 रूपये और 410 रूपये
Ans.(b)
Sol. 12.5% of profit = (12.5/100)×880=Rs.110
Remaining Rs. 770 is divided in the ratio
= 5000 : 6000 = 5 : 6
Profit to Anu = (5/11)×770+110
= Rs. 460
Profit to Bimla = (6/11)×770
=Rs.420
Required profits are Rs. 460 and Rs. 420

Q7. If [a + (1/a)]² – 2[a – (1/a)] = 12, then which of the following is a value of ‘a’?
यदि [a + (1/a)]² – 2[a – (1/a)] = 12 है,तो निम्नलिखित में से क्या ‘a’ का मान है?
(a) – 8 + √3
(b) – 8 – √3
(c) – 8 + √5
(d) None of these/इनमें से कोई नहीं

Ans.(D)
Sol.

Q8. What is the least natural number which leaves no remainder when divided by all the digits from 1 to 9?
1 से 9 तक सभी अंको से विभाजित होने पर किस न्यूनतम प्राकृतिक संख्या से कोई शेषफल प्राप्त नहीं होता है?
(a) 1800
(b) 1920
(c) 2520
(d) 5040
Ans.(c)
Sol. Required number = L.C.M. of 1, 2, 3, 4, 5, 6, 7, 8, 9
= 2 × 2 × 3 × 5 × 7 × 2 × 3 = 2520

Q9. Three sets of English, Mathematics and Science books containing 336, 240 and 96 books respectively have to be stacked in such a way that all the books are stored subject wise and the height of each stack is the same. Total number of stacks will be.
अंग्रेजी, गणित और विज्ञान की पुस्तकों के तीन सेट में क्रमशः 336, 240 और 96 पुस्तकें हैं, इन्हें इस तरह से स्टैक किया जाना है कि सभी पुस्तकें विषयवार संग्रहित हों और प्रत्येक स्टैक की ऊंचाई समान हो। स्टैक की कुल संख्या कितनी होगी?
(a) 14
(b) 21
(c) 22
(d) 48

Ans.(a)
Sol. Number of books in each stack = H.C.F. of 336, 240 and 96 = 48.
Hence, total number of stacks = ((336 + 240 + 96)/48)=672/48 = 14.

Q10. If (x/a) + (y/b) = 3 and (x/b) – (y/a) = 9, then what is the value of x/y?
यदि (x/a) + (y/b) = 3 और (x/b) – (y/a) = 9 है,तो x/y का मान कितना है?
(a) (b + 3 a)/(a – 3 b)
(b) (a + 3 b)/(b – 3 a)
(c) (1 + 3 a)/(a + 3 b)
(d) ( a + 3 b²)/(b – 3 a²)

Ans.(a)
Sol. GIVEN
(x/a)+(y/b)=3 ,&(x/b)–(y/a)=9
xb+ya=(3ab)┬ …..(1) & xa-yb=9ab……..(2)

put value of 3ab from eq 2 in eq 1
xb+ya=(xa–yb)/3
3xb + 3ya = xa – yb
3xb – xa = – yb – 3ya
x (3b – a) = – y (b + 3a)
⇒x/y=((b+3a)/(a–3b))

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