{"id":34512,"date":"2019-09-05T14:19:51","date_gmt":"2019-09-05T08:49:51","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=34512"},"modified":"2019-09-05T14:19:51","modified_gmt":"2019-09-05T08:49:51","slug":"maths-quiz-questions-for-rrb-group-d-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/maths-quiz-questions-for-rrb-group-d-pdf\/","title":{"rendered":"Maths Quiz Questions for RRB Group-D PDF"},"content":{"rendered":"

Maths Quiz Questions for RRB Group-D PDF<\/strong><\/span><\/h1>\n

Download Top-15 RRB Group-D Maths Quiz Questions PDF.\u00a0RRB GROUP-D Maths questions based on asked questions in previous exam papers very important for the Railway Group-D exam.<\/p>\n

Download Maths Quiz Questions for RRB Group-D PDF<\/a><\/p>\n\n

Download RRB Group-D Previous Papers PDF<\/a><\/p>\n

Take a RRB Group-D free mock test<\/a><\/p>\n

Question 1:\u00a0<\/b>Find the value of $tan 56^\\circ – tan 11^\\circ – tan 56^\\circ.tan 11^\\circ$<\/p>\n

a)\u00a00<\/p>\n

b)\u00a01<\/p>\n

c)\u00a0-1<\/p>\n

d)\u00a02<\/p>\n

Question 2:\u00a0<\/b>If sec \u03b8 + tan \u03b8 = -2, then find sec \u03b8 – tan \u03b8 = ?<\/p>\n

a)\u00a0-1\/2<\/p>\n

b)\u00a02<\/p>\n

c)\u00a01\/2<\/p>\n

d)\u00a0-2<\/p>\n

Question 3:\u00a0<\/b>Find the value of $1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{1}{3}}}}$.<\/p>\n

a)\u00a0$\\dfrac{18}{11}$<\/p>\n

b)\u00a0$\\dfrac{16}{11}$<\/p>\n

c)\u00a0$\\dfrac{27}{11}$<\/p>\n

d)\u00a0$\\dfrac{12}{11}$<\/p>\n

Take a free mock test for RRB Group-D<\/a><\/p>\n

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Question 4:\u00a0<\/b>If $x+\\frac{1}{x} = 2$ then find $x^8+\\frac{1}{x^8}$?<\/p>\n

a)\u00a08<\/p>\n

b)\u00a04<\/p>\n

c)\u00a02<\/p>\n

d)\u00a016<\/p>\n

Question 5:\u00a0<\/b>A bag contains Re.1 and Rs.2 coins in the ratio 5 : 8 amounting to Rs.420. Find the number of Re.2 coins.<\/p>\n

a)\u00a0150<\/p>\n

b)\u00a0160<\/p>\n

c)\u00a0120<\/p>\n

d)\u00a0100<\/p>\n

Question 6:\u00a0<\/b>The value of $122 + 345 – 3 \\div 1116 \\times 372 $ is :<\/p>\n

a)\u00a0466<\/p>\n

b)\u00a0469<\/p>\n

c)\u00a0446<\/p>\n

d)\u00a0460<\/p>\n

Question 7:\u00a0<\/b>Find the total number of factors of $18^{25} \\times 24^{32}$<\/p>\n

a)\u00a09360<\/p>\n

b)\u00a09600<\/p>\n

c)\u00a010126<\/p>\n

d)\u00a08480<\/p>\n

RRB Group D previous year papers<\/a><\/p>\n

Daily Free RRB Online Test<\/a><\/p>\n

Question 8:\u00a0<\/b>The reciprocal of sum of reciprocals of $\\dfrac{7}{11}$ and $\\dfrac{5}{8}$ is<\/p>\n

a)\u00a0$\\dfrac{111}{35}$<\/p>\n

b)\u00a0$\\dfrac{35}{111}$<\/p>\n

c)\u00a0$\\dfrac{27}{101}$<\/p>\n

d)\u00a0$\\dfrac{35}{121}$<\/p>\n

Question 9:\u00a0<\/b>Srujay made an error by multiplying a number with 5\/6 inplace of 6\/5. What is the percentage error ?<\/p>\n

a)\u00a031.55%<\/p>\n

b)\u00a030.55%<\/p>\n

c)\u00a032.55%<\/p>\n

d)\u00a033.33%<\/p>\n

Question 10:\u00a0<\/b>Find the total surface area of cuboid 40mmx30mmx15mm ?<\/p>\n

a)\u00a04000 sq.mm<\/p>\n

b)\u00a04500 sq.mm<\/p>\n

c)\u00a05000 sq.mm<\/p>\n

d)\u00a05500 sq.mm<\/p>\n

Question 11:\u00a0<\/b>The length and the breadth of the rectangle are 14m and 16m. If the cost of fencing is Rs 6 per m, then find the cost of fencing the rectangle.<\/p>\n

a)\u00a0Rs300<\/p>\n

b)\u00a0Rs180<\/p>\n

c)\u00a0Rs360<\/p>\n

d)\u00a0Rs240<\/p>\n

RRB Group-D Important Questions (download PDF)<\/a><\/p>\n\n

Question 12:\u00a0<\/b>In a rhombus, the length of the two diagonals are 8cm and 10cm respectively. Find the area of the rhombus?<\/p>\n

a)\u00a030<\/p>\n

b)\u00a020<\/p>\n

c)\u00a040<\/p>\n

d)\u00a050<\/p>\n

Question 13:\u00a0<\/b>If 1 apple, 2 bananas and 7 oranges cost Rs 34, then find the cost of 4 apples 8 bananas and 28 oranges?<\/p>\n

a)\u00a0102<\/p>\n

b)\u00a0154<\/p>\n

c)\u00a0136<\/p>\n

d)\u00a0148<\/p>\n

Question 14:\u00a0<\/b>If a = log 2 and b = log 3 and c = log 5, then find the value of log l800 in terms of a, b and c.<\/p>\n

a)\u00a02a + 2b + 3c<\/p>\n

b)\u00a0a + 2b + 3c<\/p>\n

c)\u00a03a + 2b + c<\/p>\n

d)\u00a03a + 2b + 2c<\/p>\n

Question 15:\u00a0<\/b>If three coins are tossed, what is the probability that atleast 2 heads occur?<\/p>\n

a)\u00a00.5<\/p>\n

b)\u00a00.25<\/p>\n

c)\u00a00.2<\/p>\n

d)\u00a00.1<\/p>\n

General Science Notes for RRB Exams (PDF)<\/a><\/p>\n\n

Answers & Solutions:<\/strong><\/span><\/p>\n

1)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

$56-11 = 45$
\nApplying \u2018tan\u2019 on both sides
\n$tan (56-11)^\\circ = tan 45^\\circ$ \u2192 (1)
\nWe know that $tan (A-B)^\\circ = \\dfrac{tan A^\\circ – tan B^\\circ}{1+tan A^\\circ.tan B^\\circ}$
\nThen, (1) becomes
\n$\\dfrac{tan 56^\\circ – tan 11^\\circ}{1+tan 56^\\circ.tan 11^\\circ} = 1$ (Since, $tan 45^\\circ = 1$)<\/p>\n

$tan 56^\\circ – tan 11^\\circ = 1+tan 56^\\circ.tan 11^\\circ$
\n\u21d2 $tan 56^\\circ – tan 11^\\circ – tan 56^\\circ.tan 11^\\circ = 1$<\/p>\n

2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

$sec^2 \u03b8 – tan^2 \u03b8 = 1$<\/p>\n

$(sec \u03b8 – tan \u03b8)(sec \u03b8 + tan \u03b8) = 1$<\/p>\n

$(sec \u03b8 – tan \u03b8) (-2) = 1$<\/p>\n

$(sec \u03b8 – tan \u03b8) = -1\/2$<\/p>\n

So the answer is option A.<\/p>\n

3)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

$1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{1}{3}}}}$<\/p>\n

$= 1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{1}{\\dfrac{4}{3}}}}$<\/p>\n

$= 1+\\dfrac{1}{1+\\dfrac{1}{1+\\frac{3}{4}}}$<\/p>\n

$= 1+\\dfrac{1}{1+\\dfrac{1}{\\dfrac{7}{4}}}$<\/p>\n

$= 1+\\dfrac{1}{1+\\dfrac{4}{7}}$<\/p>\n

$= 1+\\dfrac{1}{\\dfrac{11}{7}}$<\/p>\n

$= 1+\\dfrac{7}{11} = \\dfrac{18}{11}$<\/p>\n

4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

$x+\\frac{1}{x} = 2$<\/p>\n

Squaring on both sides<\/p>\n

$(x+\\frac{1}{x})^2 = 4$<\/p>\n

$x^2+\\frac{1}{x^2}+2 = 4$<\/p>\n

$x^2+\\frac{1}{x^2} = 2$<\/p>\n

Squaring on both sides<\/p>\n

$(x^2+\\frac{1}{x^2})^2 = 4$<\/p>\n

$x^4+\\frac{1}{x^4}+2 = 4$<\/p>\n

$x^4+\\frac{1}{x^4} = 2$<\/p>\n

Squaring on both sides<\/p>\n

$(x^4+\\frac{1}{x^4})^2 = 4$<\/p>\n

$x^8+\\frac{1}{x^8}+2 = 4$<\/p>\n

$x^8+\\frac{1}{x^8} = 2$<\/p>\n

So the answer is option C.<\/p>\n

5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Let the number of Re.1 coins be 5x.
\nTotal value of Re.1 coins = Rs.5x*1 = Rs.5x
\nLet the number of Rs.2 coins be 8x.
\nTotal value of Rs.2 coins = Rs.8x*2 = Rs.16x
\nTotal money in the bag = Rs.5x+Rs.16x = Rs.21x
\nGiven, Rs.21x = Rs.420
\n\u21d2 x = 20
\nHence, The number of Rs.2 coins = 8x = 8*20 = 160<\/p>\n

6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Using BODMAS rule we can write the equation as,<\/p>\n

$467 – \\frac{1}{372} \\times 372 $<\/p>\n

$467 – 1 = 466$<\/p>\n

Hence, option A is the correct answer.<\/p>\n

7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

$18^{25} \\times 24^{32} = 2^{25} \\times 3^{50} \\times 2^{96} \\times 3^{32}$
\nThis equals $2^{121} \\times 3^{82}$
\nHence, the number of factors equals $122 \\times 83 = 10126$<\/p>\n

8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Reciprocals of given numbers are $\\dfrac{11}{7}$ and $\\dfrac{8}{5}$<\/p>\n

$\\dfrac{11}{7}+\\dfrac{8}{5} = \\dfrac{55+56}{35} = \\dfrac{111}{35}$<\/p>\n

Required number = Reciprocal of $\\dfrac{111}{35} = \\dfrac{35}{111}$<\/p>\n

9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Let us assume the number to be 30 as it is lcm of 5 and 6.
\nOriginal value will be (6\/5)*30=36
\nWrong value obtained is (5\/6)*30=25
\nPercentage error =((36-25)\/36)*100
\n=30.55%<\/p>\n

10)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Total surface area of cuboid = 2(lb+bh+hl) = 2[(40)(30)+(30)(15)+(15)(40)] = 2[1200+450+600] = 2[2250] = 4500 sq.mm<\/p>\n

So the answer is option B.<\/p>\n

11)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Perimeter of rectangle = 2(14+16)m = 60m
\nTotal cost of fencing = Rs (6*60)= Rs 360<\/p>\n

12)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Area of the rhombus = 1\/2 *8 *10 = 40<\/p>\n

13)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Let the cost of 1 apple, 1 banana and 1 orange be x, y, z respectively.
\nx + 2y + 7z = 34 \u00a0 \u00a0—(a)
\nWe have to infer the value of 4x + 8y + 28z.
\nMultiplying equation (a) by 4,
\n4x+8y+28z = 34*4 = 136
\nHence, the required cost is 136.<\/p>\n

14)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

1800 = $2^3 * 3^2 * 5^2$
\n=> log 1800 = log ($2^3 * 3^2 * 5^2$)
\n=> log 1800 = log $2^3$ + log $3^2$ + log $5^2$
\n=> log 1800 = 3 log 2 + 2 log 3 + 2 log 5
\n=> log 1800 = 3a + 2b + 2c<\/p>\n

15)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

The total number of possibilities are TTT, TTH, THT, HTT, THH, HTH, HHT and HHH.
\nWe need to select those cases in which there are at least 2 Hs.
\nThere are 4 such cases out of 8.
\n=> Probability = 4\/8 = 0.5<\/p>\n\n

DOWNLOAD APP FOR RRB FREE MOCKS<\/a><\/p>\n

We hope this Maths Quiz Questions\u00a0 PDF for RRB Group-D Exam will be highly useful for your preparation.<\/p>\n","protected":false},"excerpt":{"rendered":"

Maths Quiz Questions for RRB Group-D PDF Download Top-15 RRB Group-D Maths Quiz Questions PDF.\u00a0RRB GROUP-D Maths questions based on asked questions in previous exam papers very important for the Railway Group-D exam. Download RRB Group-D Previous Papers PDF Take a RRB Group-D free mock test Question 1:\u00a0Find the value of $tan 56^\\circ – tan […]<\/p>\n","protected":false},"author":41,"featured_media":34514,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[169,125,31,1679,1601],"tags":[489,491,1637,1976,1647],"class_list":{"0":"post-34512","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-downloads","8":"category-featured","9":"category-railways","10":"category-rrb-group-d","11":"category-rrb-ntpc-je","12":"tag-railway-exam","13":"tag-rrb","14":"tag-rrb-exam","15":"tag-rrb-group-d-2019","16":"tag-rrb-mocks"},"better_featured_image":{"id":34514,"alt_text":"Maths Quiz Questions for RRB Group-D PDF","caption":"Maths Quiz Questions for RRB Group-D PDF","description":"Maths Quiz Questions for RRB Group-D 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