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निम्नलिखित संख्या श्रृंखला में प्रश्नवाचक चिन्ह (?) के स्थान पर क्या आयेगा ?
Q1. 1 7 ? 2520 85680 3855600
(1) 56
(2) 280
(3) 105
(4) 168
(5) 501
Q2 23 134 ? 689 1133 1688
(1) 365
(2) 356
(3) 456
(4) 465
(5) 366
Q3 256 137 ? 1353 12705 13176.15
(1) 226
(2) 220
(3) 230
(4) 236
(5) 240
Q 4 – 5: What value will come in place of question mark (?) in the following question?
निम्नलिखित प्रश्न में प्रश्नवाचक चिन्ह (?) के स्थान पर क्या आयेगा ?
420 × 276 + 315 × 256 + 630 × 216 + 420 × 408 = ?
(1) 5040
(2) 25600
(3) 504000
(4) 50400
(5) 540000
Q 5 36% of 450 + 48% of 520 – 68% of 750 = ?
450 का 36% + 520 का 48% – 750 का 68% = ?
(2) – 89.4
(3) 89.4
(4) – 98.4
(5) 94.8
Q 6 Simple interest on Rs.P at 10% p.a. for a certain period is Rs.200 more than the interest on Rs.(P – 1000) at 8% p.a. for the same period. Find the value of P, if time has the value equal to first composite number.
P रु. पर 10% वार्षिक ब्याज की दर से किसी निश्चित समयावधि के लिए साधारण ब्याज (P – 1000) रु. पर 8% वार्षिक ब्याज की दर से उसी निश्चित समयावधि के लिए प्राप्त साधारण ब्याज से 600 रु. अधिक है | P का मान ज्ञात कीजिये, यदि समय का मान पहली संयुक्त संख्या के बराबर है |
(1) Rs. /रु.2500
(2) Rs. /रु.2000
(3) Rs. /रु.3000
(4) Rs. /रु.3500
(5) None of these/इनमें से कोई नहीं
Q 7 The speed of a boat is thrice that of the speed of the current. The total time taken by a boatman to reach its starting point from a point 72 km farther the departure point is 10.8 hours. What is the speed of the boat in downstream?
एक नाव की चाल धारा की चाल का तीन गुना है | नाविक द्वारा 72 किमी. दूर स्थित किसी बिंदु से अपने मूल बिंदु पर वापस आने में 10.8 घंटे लगते हैं | नाव की चाल अनुप्रवाह में क्या है ?
(1) 15 km/hr. /किमी. /घंटा
(2) 20 km/hr. /किमी. /घंटा
(3) 5 km/hr. /किमी. /घंटा
(4) 10 km/hr. /किमी. /घंटा
(5) None of these/इनमें से कोई नहीं
Q 8 If the perimeter of a right angled triangle is 56 cm and area of the triangle is 84 sq. cm, then the length of the hypotenuse is
एक समकोण त्रिभुज का परिमाप 56 सेमी. है और त्रिभुज का क्षेत्रफल 84 वर्ग सेमी. है, तो कर्ण की लम्बाई है \
(1) 7 cm. /सेमी.
(2) 25 cm. /सेमी.
(3) 24 cm. /सेमी.
(4) 50 cm. /सेमी.
(5) None of these/इनमें से कोई नहीं
Q9 – 10: In the following questions, you have two equations numbered as I and II. You have to solve them and state the correct relationship between the given variables.
निम्नलिखित प्रश्नों में दो समीकरण I और II दी गयी हैं | आपको उन्हें हल करना है और दिए गए चरों में सही सम्बन्ध स्थापित करना है |
Q9 I. (p + 3) (p + 2) = 12
II. 2pq + 4p + 5q = 11
(1) p > q
(2) p < q
(3) p > q
(4) p < q
(5) If/यदि p = q or no relation can be established./या कोई सम्बन्ध स्थापित नहीं किया जा सकता है |
Q.10 I. p2 + 20p + 4 = 50 – 25p
II. q2 – 10q – 24 = 0
(1) p > q
(2) p < q
(3) p > q
(4) p < q
(5) If/यदि p = q or no relation can be established/या कोई सम्बन्ध स्थापित नहीं किया जा सकता है |
ANSWERS:-
Q.1 (3)
Sol. × 7, × (7 + 8), × (7 + 8 + 9), (7 + 8 + 9 + 10), × (7 + 8 + 9 +10 + 11)
Q.2 (2)
Sol. + 111, + 222, + 333, + 444, + 555
Q.3 (4)
Sol. × 0.5 + 9, × 1 + 99, × 1.5 + 999, × 2 + 9999, × 2.5 + 99999
Q.4 (3)
Sol. 420 × 276 + 315 × 256 + 630 × 216 + 420 × 408 = ?
? = 5040 × 23 + 5040 × 16 + 5040 × 27 + 5040 × 34
? = 5040 × 100
? = 504000
Q.5 (4)
Sol. 36% of 450 + 48% of 520 – 68% of 750 = ?
450 का 36% + 520 का 48% – 750 का 68% = ?
162 + 249.6 – 510 = ?
? = – 98.4
Q.6 (4)
Sol.
40P – 32P = 60000 – 32000
8P = 28000
P = 3500
Q.7 (2)
Sol. According to the question/प्रश्नानुसार,
x = 5
Speed downstream/अनुप्रवाह चाल = 15 + 5 = 20 km/hr. /किमी. /घंटा
Q.8 (2)
Sol. We know that/हम जानते हैं कि,
× base/आधार × height/ऊँचाई = 84 … (I)base/आधार × height/ऊँचाई = 168
and/और, Base/ऊँचाई + Height/ऊँचाई + Hypotenuse/ऊँचाई = 56 … (II)
Now, taking all the possible multiples of 168 which satisfies (II) and also Pythagoras theorem, we get
अब, 168 के सभी संभव गुणजों को लेकर जोकि समीकरण (II) को और पाइथागोरस प्रमेय को संतुष्ट करते हैं, हम प्राप्त करते हैं
Hypotenuse/विकर्ण = 25 cm./सेमी.
Q.9 (5)
Sol. I. (p + 3) (p + 2) = 12
p2 + 5p + 6 = 12
p2 + 5p – 12 = 0
p2 + 6p – p – 12 = 0
p(p + 6) – 1(p + 6) = 0
p = – 6, 1
II. 2pq + 4p + 5q = 11
q = – 5, 1
Hence, no relation can be established./ अतः, कोई सम्बन्ध स्थापित नहीं किया जा सकता है |
Q.10 (5)
Sol. I. p2 + 20p + 4 = 50 – 25p
p2 + 45p - 46 = 0
p2 – 46p + p – 46 = 0
p(p – 46) + 1(p – 46) = 0
p = 46, – 1
II. q2 – 10q – 24 = 0
q2 – 12q + 2q – 24 = 0
(q – 12)q + 2(q – 12) = 0
(q – 12) (q + 2) = 0
q = – 2, 12
Hence, no relation can be established.
अतः, कोई सम्बन्ध स्थापित नहीं किया जा सकता है |
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