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## Advance Maths Series for SSC CGL- Mensuration Day 2

Advance Maths Series for SSC CGL- Mensuration Day 2

Dear Students, we have started a new Series for SSC CGL-Mensuration. This will cover all the important questions of Mensuration. We have made a collection of 500 most important questions of Mensuration and in this series we will be covering all these 500 questions. We will be covering 10 questions everyday form this Topic. This is the second article in this series and we have covered 10 very important questions in this article.

Advance Maths has become one of the most important part of the SSC CGL examination. We have seen an increase in the number of questions from this section year after year. This is also one of the toughest part of Maths preparation. This section not only require a firm understanding of the basics but also lots of practice to succeed in the examination. Advance Maths can be broadly divided into four parts: Mensuration, Algebra, Geometry and Trigonometry. Mensuration is a very important part and we will be covering all the important questions through this series.

Questions:

Q11. The perimeter of a rectangle and a square are 160 m each. The area of the rectangle is less than that of the square by 100 sq m. The length of the rectangle is :

आयत तथा वर्ग का परिमाप 160 मी है यदि आयत का क्षेत्रफल वर्ग के क्षेत्रफल से 100 वर्ग मी कम है तो आयत की लम्बाई कितनी है।

(a) 30 m / 30 मी

(b) 60 m / 60 मी

(c) 40 m / 40 मी

(d) 50 m / 50 मी

Q12. Perimeter of rectangular field is 160 metres and the difference between its two adjacent sides is 48 metres. The side of a square field, having the same area as that of the rectangle is:

किसी आयतकार मैदान का परिमाप 160 मी और इसकी दो  आसन्न भुजाओं का अन्तर 48 मी है । वर्ग की भुजा  क्या होगी यदि वर्ग का क्षेत्रफल आयत के क्षेत्रफल के समान है ?

(a) 32 metres / 32 मी

(b) 8 metres / 8 मी

(c) 4 metres  / 4 मी

(d) 16 metres / 16 मी

Q13. The curved surface area of a cylindrical pillar is 264 m2 and its volume is 924 m3. Taking π= 227, find the ratio of its diameter to its height.

यदि एक बेलनाकार खम्बे की वक्र पृष्ठ का क्षेत्रफल 264 मी2 और उसका आयतन 924 मी3  है तो उसके व्यास तथा लम्बाई का अनुपात होगा। मान ले π= 227 है।

(a) 7: 6

(b) 6: 7

(c) 3: 7

(d) 7: 3

Q14. The perimeters of five squares are 24 cm, 32 cm, 40 cm, 76 cm and 80 cm respectively. The perimeter of another square equal in area to sum of the areas of these squares is:

यदि पांच वर्गो का परिमाप 24 सेमी, 32 सेमी ,40 सेमी , 76 सेमी  और 80 सेमी है एक अन्य वर्ग का परिमाप पांचो वर्गो के क्षेत्रफलो के योग बराबर है तो अन्य वर्ग का परिमाप है।

(a) 31 cm  / 31 सेमी

(b) 62 cm / 62 सेमी

(c) 124 cm / 124 सेमी

(d) 961 cm / 961 सेमी

Q15. A cuboidal water tank has 216 litres of water. Its depth is 13 of its length and breadth is 12 of 13 of the difference of length and breadth. The length of the tank is:

एक घनाकार पानी की टंकी  216 लीटर पानी धारण कर सकती यदि उसकी गहराई  उसकी लम्बाई का 13 औऱ उसकी चौङाई उसकी लंबाई तथा गहराई के अन्तर के 12 का  1 3है तो उसकी लम्बाई क्या होगी ?

(a) 72 dm/72 डेसी मी

(b)18 dm/ 18 डेसी मी

(c) 6 dm/ 6 डेसी मी

(d) 2 dm/ 2 डेसी मी

Q16. The area of a rhombus is 150 cm2. The length of one of its diagonals is 10cm. The length of the other diagonal is:

एक समचतुर्भुज का क्षेत्रफल 150 सेमी2  है, इसके एक विकर्ण की लम्बाई 10 सेमी है तो दूसरे विकर्ण की लम्बाई क्या है?

(a) 25cm / 25 सेमी

(b) 30cm / 30 सेमी

(c) 35cm / 35 सेमी

(d) 40cm / 40 सेमी

Q17. The cost of carpeting a room is Rs. 120. If the width had been 4 metres less, the cost of the carpet would have been Rs. 20 less. The width of the room is:

किसी कमरे में कालीन बिछाने का खर्च 120 रुपये है यदि उसकी चौड़ाई 4 मी कम है तो क़ालीन की कीमत 20 रुपये कम हो जाती है तो इसकी चौड़ाई होगी।

(a) 24m / 24 मी

(b) 20m / 20 मी

(c) 25m / 25 मी

(d) 18.5m / 18.5 मी

Q18. The area of a field in the shape of a trapezium measures 1440 m2. The perpendicular distance between its parallel sides is 24m. If the ratio of the parallel sides is 5: 3, the length of the longer parallel side is:

एक समचतुर्भुजआकार का खेत जिसका क्षेत्रफल 1440 मी2 है  यदि उसकी समान्तर भुजाओंं  की दूरी 24 मी है और इसकी समान्तर भुजाओंं का अनुपात 5: 3 है तब लम्बी समान्तर भुजा की लम्बाई होगी।

(a) 75m / 75 मी

(b) 45m / 45 मी

(c) 120m / 120 मी

(d) 60m / 60 मी

Q19. The perimeter of a rectangle is 160 metre and the difference of two sides is 48 metre. Find the side of a square whose area is equal to the area of this rectangle?

एक आयताकार का परिमाप 160 मी है और इसकी दो भुजाओं का अन्तर 48 मी है । उस वर्ग की भुजा क्या होगी जिसका क्षेत्रफल आयत के क्षेत्रफल के बराबर है।

(a) 32 m / 32 मी

(b) 8 m / 8 मी

(c) 4 m / 4 मी

(d) 16 m / 16 मी

Q20. A circular wire of diameter 42 cm is bent in the form of rectangle whose sides are in the ratio 6 : 5. The area of the rectangle is (user π= 227)

एक वृताकार तार जिसका व्यास 42 सेमी है को आयताकार रूप में मोङा गया है जिसका भुजाओंं का अनुपात 6 : 5 है तो आयताकार का क्षेत्रफल होगा?

(a) 540 cm2 /540 सेमी2

(b) 1080 cm2 /1080 सेमी2

(c) 2160 cm2 / 2160 सेमी2

(d) 4320 cm2 / 4320 सेमी2

Solutions:

11. (d) Let the length and breadth of rectangle be x and y m. respectively.;

According to the question, 2(x + y) = 160; → x + y = 1602=80 m……..(i);

Perimeter of square = 160 m; Side of square = 1604=40m;

Now, Area of rectangle = xy; Area of square = 40 ×40=1600 m2;

Then,1600 – xy = 100; → xy = 1600 – 100 = 1500 ……… (ii);

Now, (x – y)2 = (x + y)2 – 4xy = (80)2 – 4 × 1500 = 6400 – 6000 = 400; → x – y = 400=20……..(iii);

From equations (i) and (iii), 2x = 100; → x = 1002=50 m.

12. (a) Let the length and breadth of rectangle be x and y metres respectively, Then, 2 (x + y) = 160; → x + y = 80 ………… (i);

And x – y = 48 ……….. (ii);

Adding equations (i) and (ii); 2x = 128; → x = ×1282=64;

From equation (i) y = 80 – 64 = 16; Area of rectangle = Area of square = 64 ×16 sq.m.; Side of square = 64×16 = 8 ×4=32 m.

13. (d) If r be radius of base and h the height, then ;

Curved surface of cylindrical pillar = 2πrh and volume= πr2h.; 2πrh=264 m2 …… (i); πr2h. = 924 m3 …… (ii);

On dividing (ii) by (i), we get  πr2h2πrh= 924264 m.; → r2= 924264 m.; → r = 924×2264 m=7m;

Diameter = 2 ×7=14m; From (i), H = 264π×d= 264×722×14=6m; Required ratio = 146 i.e, 7 :3

14. (c) Sides of the squares are 6 cm, 8 cm, 10 cm, 19 cm and 20 cm respectively.;

Sum of their areas = (62 + 82 +102+ 192 + 202) cm^2; = (36+64+100+361+400)cm^2 = 961 cm^2; Its side = 961=31 cm;

Required       perimeter = 4 ×31=124 cm.

15. (b) Let the length of tank = x dm ; Depth = x/3 dm; Breadth = (x – x/3 ) ×1/3 ×1/2= 2x/3 ×1/3 ×1/2= x/9 ; Volume of tank = x ×x/9 ×x/3= x^3/27 ;

According to the question, x^3/27 = 216; x^3 = 27 ×216 ; → x = (27 ×216)1/3=3×6=18 dm

16. (b) If d1 , d2 be the diagonals of a rhombus, Area = 12 d1 * d2; → 150 = 12 ×10× d2; → d2 = 1505=30 cm.

1. (a) (Let the cost of carpeting per sq. meter be Re.1.; ∴ Area of the room = 120 sq. meter;

Let the breadth of the room be x metres.; ∴ Length of room = 120x sq.metre; New cost = 120 – 20= Rs. 100;

Breadth = (x – 4 ) = 100 ; Then, 120x×(x-4)= 100; →6x (x-4)=5; →6x-24=5x= →x=24; ∷Breadth of the room=24 metres

18. (a) Let the parallel sides be 5x and 3x metres.;

Area of trapezium = 12 (sum of parallel sides) x distance between them; → 1140 = 12 (5x+3x)×24; → 12 ×8x=1440; → x = 144012×8=15;

The longer parallel side= 5x = 5 ×15=75 metres

19. (a) Let the length and breadth of rectangle are a and b respectively.;

According to the question, 2 (a + b) = 160; → a + b = 80 ………… (i); a – b = 48 …………. (ii); (On adding) 2a = 128 ; → a = 1282=64m; From equation (i),b = 80 – 64 = 16 m;

Area of rectangle = 64 ×16m2 ; Area of square = 64 ×16m2; → (Side)2 = 64 ×16; → side = 8×4=32m

20. (b) Length of wire = 227 ×42=132 cm.;

Let the length of rectangle = 6x and breadth = 5x cm; 2 (6x + 5x) = 132; → 22x = 132 → x = 13222=6; Area of rectangle = 36 ×30=1080 cm2

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