# Advance Maths Series for SSC CGL- Mensuration Day 1

## Advance Maths Series for SSC CGL- Mensuration Day 1

Advance Maths Series for SSC CGL- Mensuration Day 1

Hello friends, we are starting a new series for Maths. This will focus on the advance maths section of the SSC CGL examination. We have made a collection of 500 most important questions for Mensuration, Algebra, Trigonometry and Geometry each. In this series we will be covering all these 500 most important questions for every topic. We will cover 10 questions every day from each topic. This is the first article in this series and we have covered first 10 questions from Mensuration 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. Our main focus is to make it easier for the students to master this part.

With this series we plan to cover 500 most important questions which have the highest probability of being asked in the examination. We recommend students to solve all the questions posted in this series.

Questions:

Q1. A cistern of capacity 8000 litres measures externally 3.3 m by 2.6 m by 1.1 m and its walls are 5 cm thick. The thickness of the bottom surface is:

एक टंकी जिसकी क्षमता 8000 लीटर है । टंकी का बाहरी माप  3.3 मी × 2.6 मी × 1.1 मी है और दिवार 5 सेमी मोटी है तो टंकी की निचली सतह  की मोटाई क्या है?

(a) 1 m / 1 मी

(b) 1.1 m / 1.1 मी

(c) 1 dm / 1 डेसी मी

(d) 90 cm / 90 सेमी

Q2. The area of a circular garden is 2464 sq.m., how much distance will have to be covered if you want to cross the garden along its diameter ? (Use π= 227)

एक वृताकार पार्क का क्षेत्रफल 2464 वर्ग मी है। यदि आप पार्क के व्यास के साथ इसे पार करना चाहते है तो बताओ आपको कितनी दूरी तय करनी होगी ?

(a) 56 m / 56 मी

(b) 48 m / 48 मी

(c) 28 m 28 मी

(d) 24 m / 24 मी

Q3. If a right circular cone of height 24 cm has a volume of 1232 cm3, then the area of its curved surface (taking π= 227) is:

यदि एक लंब वृत्‍तीय शंकु जिसकी ऊंचाई 24 सेमी है तथा आयतन 1232 सेमी3 है तो बताओ की इसका वक्र पृष्ठ का क्षेत्रफल क्या होगा?

(a) 1254 cm2  / 1254 सेमी2

(b) 704 cm2 / 704 सेमी2

(c) 550 cm2 / 550 सेमी2

(d) 154 cm2 / 154 सेमी2

Q4. If the area of a triangle is 1176 cm2 and ratio of base to corresponding altitude is 3:4, then the length of this altitude of the triangle is:

किसी त्रिभुज का क्षेत्रफल 1176 सेमी2 है और उसके आधार तथा संबंधित ऊंचाई का अनुपात 3: 4  है तब त्रिभुज के इस  शीर्ष-लंब की लम्बाई कितनी है?

(a) 42 cm / 42 सेमी

(b) 52 cm / 52 सेमी

(c) 54 cm / 54 सेमी

(d) 56 cm / 56 सेमी

Q5. If the ratio of areas of two squares is 225: 256, then the ratio of their perimeters is:

यदि दो वर्गो के क्षेत्रफल का अनुपात 225: 256 है तब उनके परिमाप में अनुपात क्या होगा।

(a) 225: 256

(b) 256: 225

(c) 15: 16

(d) 16: 15

Q6. The area of an equilateral triangle is 4003 sq.m. Its perimeter is:

किसी समबाहु त्रिभुज  का क्षेत्रफल 4003 सेमी है तो उसका परिमाप कितना है ?

(a) 120 m/ 120 मी

(b) 150 m / 150 मी

(c) 90 m / 90 मी

(d) 135 m / 135 मी

Q7. The curved surface area of a cylindrical pillar is 264 m2 and its volume is 924 m3. The ratio of its diameter to its height is [user π= 227]

एक बेलनाकार स्तंभ का वक्र पृष्ठीय क्षेत्रफल 264 मी2  है और इसका आयतन 924 मी3 है तो इसके व्यास तथा ऊंचाई का अनुपात क्या है ?

(a) 7 : 6

(b) 6 : 7

(c) 3 : 7

(d) 7 : 3

Q8. There is a rectangular tank of length 180 m and breadth 120 m in a circular field. If the area of the land portion of the field not covered by the tank is 40000 m2, what is the radius of the field? (Take π= 227)

एक आयताकार टैकं जिसकी लम्बाई 180 मी और चौड़ाई 120 मी है  एक वृताकार खेत में स्थित है यदि उस खेत के शेष-भाग का क्षेत्रफल 40000 मी2 है तो उस खेत की त्रिज्या की लम्बाई होगी ? मान ले (π= 227) है

(a) 130m / 130 मी

(b) 135m / 135 सेमी

(c) 140m / 140 सेमी

(d) 145m /145 मी

Q9. A cuboidal water tank contains 216 litres of water. Its depth is 13 of its length and breadth is 12 of 1 3 of the difference between length and depth. The length of the tank is:

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

Q10. The area of a triangle is 216 cm2 and its sides are in the ratio 3 : 4 : 5. The perimeter of the triangle is:

किसी  त्रिभुज का क्षेत्रफल 216 सेमी2 है और इसकी भुजाओं का अनुपात  3 : 4 : 5 है तो  त्रिभुज का परिमाप है।

(a) 6cm / 6 सेमी

(b) 12 cm / 12 सेमी

(c) 36 cm / 36 सेमी

(d) 72 cm / 72 सेमी

##### Solutions:
1. (c) Walls are 5 cm thick.; Internal length = (330 – 2 ×5) cm.=320 cm.; Breadth = (260 – 10) cm. = 250 cm.; Height = (110 – x) cm.; Here, the cistern is assumed to be open and x is the thickness of bottom.; 320 ×250 ×(110- x)=8000 litres ; → 320 ×250 ×(110- x)=8000× 1000 cm3 ; → (110- x)= 8000000320×250 ; → 110 – x = 100; → x = 10 cm. = 1 dm.
2. (a) πr2=2464 sq.m.; → r2= 2464×722=784; → r = 28 m.; Required distance = 2r = 2 ×28=56 metres
3. (c) Volume of the cone = 13 πr2h; → 1232 = 13 ×227 ×r2 ×24; → r2= 1232×3×722×24=49; r = 7                        Slant height = (24)2+ (7)2 = 576+49= 625=25 cm.; Area of curved surface = πrl= 227 ×7×25=550 cm2
1. (d) Let the base and altitude be 3x and 4x respectively; According to question,;  12 base × altitude=1176cm2 ; Or, 12 ×3x ×4x=1176; 12x2 = 1176 ×2; X2 = 1176×212 or, x2 = 196; x = 196=14cm.; Altitude of a triangle = 4x; = 4 ×14cm=56cm
2. (c) Ratio of area = 225256 Ratio of sides = 225256= 1516 ; Ratio of perimeters= 4×154×16= 1516=15 :16
3. (a) Area of equilateral triangle = 34 ×(Side)2; →  34 ×(Side)2= 4003; → (Side)2=4003 ×43 ; Side = 4×400=40 metres ; Perimeter = 3 ×40=120 metres
4. (d) Let the radius of base of cylindrical pillar be r and height be h cm. Then, 2πrh=264……….(i); And πr2h=924……..(ii); On dividing equation (ii) by (i), πr2h2πrh= 924264 ; → r = 2×924264=7; From equation (i),2π×7×h=264 ; →h= 264×72×22×7=6;   Required ratio = 2 ×7 :6=7 :3
5. (c) Area of the tank = 180 ×120=21600 m2 ; Total area of the circular plot = 40000 + 21600 m2 ; πr2=61600; → r2 = 61600×722=2800×7; → r = 2800×7=2×7×10=140 m
6. (b) Let the length of the tank be x cm.; Depth = x3  ; Breadth = 12 ×13 ×(x- x3)= x9 ; Now, X ×x3×x9=216×1000; → x3 = 27×216×1000; → x = (27×216×100)1/3; → x = 3 ×6×10=180 cm=18 dm
7. (d) Let the sides be 3x, 4x and 5x respectively.; Here, (3x)2 + (4x)2 = (5x)2 ; Hence, the triangle is right          angled.; 12 ×3x ×4x=216; → 6x2 = 216 → x2 = 2166=36; x = 36=6; Perimeter of triangle = (3x + 4x + 5x) cm = 12x cm= 12 ×6=72 cm.

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